lecture videos for scalars & vectors
scalars & vectors video#1 Review of the basic differences
between vectors and scalars. Also adding vectors in one dimension.
scalars & vectors video#2 Review of how to find the
magnitude of the resultant of vector addition in two dimentions using
tip-to-tail method and the parallelogram method.
scalars & vectors video#3 Solving vector problems
graphically using a protractor and a ruler.
scalars & vectors video#4 Review of trig function that
you can use to solve vector problems. Sine, Cosine, and Tangent basics.
scalars & vectors video#5 Practice vector problems using
trig functions as well as quicknotes on superposition and multiplying vectors
by scalars.
i will upload a bunch of videos if you are really entrusted in learning
physics.
Physics
XI Numericals Karachi Board
Section A
1. If one of the rectangular components of a
force of 100 N is 50 N, find the other component.
2. Two bodies A and B are attached to the ends
of a string passing over a frictionless pulley such that the masses hang
vertically. If the mass of one body is 96kg, find the mass of second body,
which moves downward with an acceleration of 0.2 m/s2 and the tension in the
string. (g = 9.8 m/s2)
3. An artillery cannon is pointed upward at an
angle 35° with respect of horizontal and fires a projectile with an initial
velocity of 200m/s. If the air resistance is negligible, find the maximum
height reached by the projectile and the range of the projectile.
4. Two vectors have magnitudes 4 and 5 units.
The angle between them is 30°. Taking the first vector along x-axis, calculate
the magnitude and the direction of the resultant.
5. Given that , find , , and the angle between.
6. A 100 gm bullet is fired into a 12kg block
which is suspended by a long cord. If the bullet is embedded in the block and
the block and the block rises by 5cm. What was the speed of the bullet?
7. The radius of the moon is 27% of the earth’s
radius and its mass is 1.2% of the earth’s mass. Find the acceleration due to
gravity on the surface of the moon. How much does a 980 N body weigh there.
8. Find the value of P for which the following
vectors are perpendicular to each other.
9. A boy throws a ball vertically upward with a speed
of 25m/s. On the way down, it is caught at a point 5m above the ground. How
fast was it coming down at this point? How long did the trip take?
10. A string 1m long would break when its
tension is 69.6N. Find the greatest speed at which a ball of mass 2kg can be
whirled with the string in a vertical circle.
11. A ladder rests against a smooth wall at an
angle of 60° with the ground. The ladder weighs 200 N and its centre of gravity
is at one thirds of its length from the base. Determine the frictional force,
which prevents the ladder from slipping, and the coefficient of static
friction.
12. If two vectors are such that A = 3 and B = 2
and = 4, evaluate and .
13. If find a unit vector parallel to .
14. A mortar shell is fired at ground level
target 490m away with an initial velocity of 98m/s. Find the two possible
values of the launch angle. Calculate the minimum time to hit the target.
15. Find how deep from the surface of earth a
point where earth a point is where the acceleration due to gravity is half the
value on the earth’s surface.
16. Given . Find the magnitude of vectors , and
.
17. A minibus starts moving from the position of
rest at a bust stop with a uniform acceleration during the 10th minute of its
motion it covers a distance of 95m. Calculate its acceleration and the total
distance it covers in 10 minutes.
18. Calculate the work done by a force given by
in displacing a body from the position A to the position B. The position
vectors A and B are
Section B
1. A body of 0.5kg attached to a spring is
displaced from its equilibrium position and is released. If spring constant K
is 50N/m, find the time period and the frequency.
2. A car has siren sounding 2KHz tone. What
frequency will be detected by stationary listener as the car is approaching him
at 80km/h. (speed of sound = 1200 km/h).
3. Calculate the speed of sound in air at
atmospheric pressure P = 1.01 x 105 N/m2, taking g = 1.40 and r = 1.2 kg/m3.
4. Calculate the length of second’s Pendulum at
a place where g = 10.0 m/s2.
5. When mass m is hung on a vertical spring, it
stretches through 6cm. Determine its period of vibration if its is pulled down
a little and released.
6. A guitar string has a linear density of 7.16
gm/m and is under tension of 152 N. the fixed supports of the string are 89.4
cm apart. If its vibrates in three segments. Calculate the speed, the
wavelength and the frequency of the standing wave.
7. Calculate the length of second’s pendulum on
the surface of moon where the acceleration due to gravity is 0.167 times that
on the earth’s surface.
8. A source of sound and a listener are moving
towards each other with velocities, which are 0.5 time, and 0.2 time the speed
of sound respectively. If the source is emitting 2KHz tone. Calculate the
frequency heard by the listener.
9. A stationary wave is set in a 1.5 metre long
string fixed at both ends. The string vibrates in five segments when driven by
a frequency of 100 Hz. Calculate the wavelength and the fundamental frequency.
Section C
1. How many fringes will pass a reference
point if the movable mirrors of the Michelson’s Interferometer are moved by
0.08 mm? The wavelength of light used is 5800 A°.
2. A green light of wavelength 5400 A° is
diffracted by a diffraction grating having 2000 lines/cm. Compute the angular
deviation of the third order image.
3. Find the distance at which an object should
be placed in front of a convex lens of focal length 20 cm to obtain an image of
double its size.
4. If the radius of 14th ring is 1mm and the
radius of curvature of the lens 126 mm. calculate the wavelength of light.
5. Red light falls normally on a diffraction
grating ruled 4000 lines /cm and the second order image is diffracted 34° from
the normal. Compute the wavelength of red light in angstroms.
6. In a compound microscope, the focal lengths
of the objective and eyepiece are 0.8cm and 2.5cm respectively. The real image
is formed by the objective is 16cm from it. Determine the Magnifying power if
the eyepiece is held close to the eyepiece and the image is formed 25cm from
the eyepiece.
7. Interference fringes were produced by two
slits on a screen 0.8m from them when the light of wavelength 5.8 x 10m was
used. If the separation between the first and fifth bright fringe is 2.5 mm.
Calculate the separation of the two slits.
8. If the radius of 10th ring is 0.5mm when the
light o 6.00 x 10-7 m is used. What is the radius of curvature of the lens
used?
9. The length of a compound microscope is 30cm.
The focal length of the objective is 0.25 cm and that of the eyepiece is 10cm.
Calculate its magnifying power if the final image is formed at a distance of
250mm from the eyepiece.
Numericals of Physics 1st Year (XI) Karachi
Board
1st Year
Physics True and False
The Scope of Physics
1. Screw and liver was invented by Newton.
2. Pythagouras was famous in mathematics.
3. Logarathim was invented by Al-Beruni.
4. Omer Khyyam was mathematician and he was also
a poet.
5. Pin hole camera was invented by Al-Razi.
6. Ibn-e-Sina was famous for his work in
chemistry.
7. Al-Razi wrote about 200 books.
8. 20th century is the century of Physics.
9. Dimension of volume is L3.
10. Dimension of Linear momentum is MLT-1.
11. 9.8 contain two significant figures.
12. In C.G.S system the fundamental units of
length, mass and time are kg, km and hour.
13. In British Engineering system the unit of
force, length and time are chosen as the fundamental unit.
14. In M.K.S system unit of mass is pound.
15. Candela is the unit of luminous intensity.
16. The branch of physical science, which deal
with interaction of matter and energy, is called physics.
17. The biological science deals with non-living
things.
18. The class of science, which deals with the
properties, and behaviour of living things is called physical science.
19. Ibn-e-Sina was famous in the field of
mathematics.
20. In C.G.S system the unit of force is pound.
21. In the field of research the strong
incentive comes from Bible.
22. The new era of modern physics began near the
end of 19th century.
23. Al-Khwarizmi was the founder of analytical
algebra.
24. Egyptian for the first time manufactured
paper.
25. Chinese used to measure the flood level in
the river Nile.
26. Ibn-e-Sina was a great astronomer.
27. The author of Alsh-Shifa was Al-Beruni.
28. The author of Al-Qannun-Fil-Tib was
Ibn-e-Sina.
29. Yakoob- Bin Ishag Al-Kindi wrote many books
on the mathematics, astronomy, medicine and other subjects.
30. In atomic clock the time be measured to an
accuracy of part in 1012.
31. The unit of electric current is coulomb.
32. The SI unit of thermodynamic temperature is
celsius.
33. The dimension of area is L-2.
34. The dimension of force is MLT-2.
35. 14.71 has four significant figures.
1st Year
Physics Multiple Choice Questions
The Scope of Physics
1. The branch of physical science, which deals
with interaction of matter and energy, is called __________.
(Physics, Chemistry, Biology)
2. The new era of modern physics began near the
end of __________.
(17th century, 18th century, 19th century)
3. Screw and lever were invented by __________.
(Newton, Huygen, Archimedes)
4. Phythagoras is famous in __________.
(Physics, Chemistry, None of these)
5. In the field of research the strong incentive
comes from __________.
(Bible, Quran, Ingeel)
6. Number of ayah which are taken from Surah
Nooh for our book are __________.
(11 and 12, 13 and 14, 15 and 16)
7. Number of ayah taken from Surah ‘Al Imran’
__________.
(170 and171, 180 and 181, 190 and 191)
8. Al-Khawarizmi was the founder of __________.
(Microbiology, Analytical Algebra, Physics)
9. Logarithm was invented by __________.
(Al- Beruni, Al-Khawarizmi, Ibn-e- Sina)
10. In Muslim world the man was both a poet and
a mathematician is __________.
(Omer Khyyam, Al-Khawarizmi, Al-Beruni)
11. Kitabul Manazir was written by __________.
(Yaqoob Bin Ishaq, Ibn-e-Sina, Ibn-al-Haitham)
12. Pin hole camera was invented by __________.
(Ibn-al-Haithan, Al-Razi, Al-Beruni)
13. Ibn-e- Sina was famous for his research in
the field of __________.
(Medicine, mathematics, physics)
14. Muslim scientist who wrote about 200 books
is __________.
(Abn-e-Sina, Al-Razi, Omer khyyam)
15. 20th century is called the century of
__________.
(Physics, Chemistry, Mathematics)
16. Dimension of acceleration is __________.
(LT-1, LT-2, L-1T)
17. The significant figures of 16, 7 are
__________.
(7, 6 and 7, 1, 6 and 7)
18. The author of Kitab-ul-Masoodi was
__________.
(Al-Beruni, Ibn-e-Sina, Ibn-al-Haitham)
19. The author of Al-Qanun-Fil-Tib was
__________.
(Al-Beruni, Ibn-e-Sina, Ibn-al-Haitham)
20. Alsh-Shifa an encyclopedia of philosophy was
written by __________.
(Al-Beruni, Ibn-e-Sina, Abn-al-Haitham)
21. Atomic clock is a (briefly) radio
transmitter giving out short waves of wavelength about __________.
(3cm, 3m, 3A°)
22. The time interval occupied 9192631770 cycles
of a specified energy change in the Cesium atom is taken as equal to one
__________.
(second, minute, hour)
23. The ampere is the unit of __________.
(time, electric current)
24. Mole is the amount of substance of a system
which contains as many elementary entities as there are atom in 0.012 kg of
__________.
(Cesium – 133, Uranium – 298, Carbon – 12)
25. The dimension of volume is __________.
(L2, L-2, L3)
26. The dimension of velocity is __________.
(LT-2, L-1L2, LT-1)
27. The dimension of linear momentum is
__________.
(MLT-1, ML-1T, M-1LT)
28. The number of 6408.2 has __________
significant figure(s).
(one, four, five)
29. The circumference of a circle of radius 3.5
cm is __________.
(21.99cm, 38.49 cm,179.62 cm)
30. The volume of a sphere of radius 3.5 cm is
__________.
(21.99 cm3, 38.49cm3, 179.62cm3)
31. Al Khawarizmi was the founder of __________.
(Decimal system, Geomtery, Analytical Algebra)
32. A number, which is reasonably reliable, is
called __________.
(Ratio, Function, Significant Figure)
33. Electromagnetic wave theory of light is
proposed by __________.
(Maxwell, Newton, Huygen)
34. Wave mechanics were introduced by __________.
(De-Broglie, Maxwell, Newton)
35. Natural Radioactivity was discovered by
__________.
(Madam Curie, Bacquerel, Max-Plank)
1st Year
Physics Fill in the Blanks
Geometrical Optics
1. A lens is a piece of __________ material that
can focus a transmitted beam of light.
2. A lens is a piece of transparent material
that can focus a __________ beam of light.
3. A lens is usually bounded by two __________
surfaces.
4. Lenses fall into __________ categories.
5. A convex lens is thicker in the middle and
__________ on the edges.
6. A convex lens is __________ in the middle and
thinner on the edges.
7. A convex lens __________ the light rays
towards its optical axis.
8. A convex lens converges the light rays
towards its __________.
9. A convex lens is thinner in the middle and
__________ on the edges.
10. A diverging lens is __________ in the middle
and thicker on the edges.
11. A concave lens __________ the light rays
from its optical axis.
12. A diverging lens bends light rays from its
__________.
13. The point to which the light rays are
brought to focus is called __________.
14. The point to which the light rays are
brought to __________ is called principal focus.
15. The distance between the optical centre of
the lens and its principle focus is called __________.
16. The distance between the __________ of the
lens and its principle focus is called its focal length.
17. The point is the lens through which the
light rays will pass without any deviation is called its __________.
18. The point is the lens through which the
light rays will pass without any deviation is called its __________.
19. In Convex lens, when the object is placed in
between F and 2F then its image will form __________ on the other side of the
lens.
20. Conventionally the focal length of the
diverging lens is taken as __________.
21. When the two lenses are combined to form a
single lens, then this lens is called __________ lens.
22. The defect in the lens is due to the fact
that the focal points of the light rays far from the optical axis of a
spherical lens are different from those rays passing through the centre is
called __________.
23. Chromatic aberration can be reduced to a
greater extent by the combination of __________ lenses.
24. The angle subtended by the object at the eye
is called __________.
25. The least distance up to which a normal
person could see the object without taking any strain on his eyes is called
least distance of __________.
26. Compound microscope is an optical device,
which is used to see the __________ object with very high magnification.
27. In compound microscope an eyepiece of
__________ focal length is used.
28. In compound microscope to obtain the large
magnification lenses of __________ focal length are used.
29. In compound microscope the eyepieces will be
used as a __________.
30. Telescope is used to see __________ object.
31. The focal length of the objective in
astronomical telescope is __________ than the focal length of eyepiece.
32. Galilean telescope is used to see the object
on __________.
33. In Galilean telescope the convex lens will
be used as an __________.
34. Terrestrial telescope consists of __________
lenses.
35. In spectromter, collimeter produces a
__________ beam of light.
36. The spectrum of light in which the images
overlap each other is called __________ spectrum.
37. The front of eye is covered by a transparent
membrane called __________.
38. The abnormality in which the image of the
distant object is focused in from of retina is called __________.
39. In the case of the abnormality named as
Myopia the person is said to __________.
40. Short Sightedness can be corrected by using
a __________ a lens.
The Scope of Physics
1. Physical sciences and biological sciences are
the two branches of __________.
2. The branch of physics, which deals with the
interaction of matter and energy, is called __________.
3. Ibn-e-Sina was famous in the field of
__________.
4. Medical text book Al-Qannun-Fil-Tib was
written by __________.
5. In C.G.S system the unit of force is
__________.
6. Unit of luminous intensity is __________.
7. Dimension of volume is __________.
8. Dimension of linear momentum is __________.
9. 16.7 contain __________ significant figures.
10. Logarithm was invented by __________.
11. The biological science deals with
__________.
12. The class of science, which deals with the
properties and behaviour of non-living matter is called __________ science.
13. Chinese for the first time manufactured __________.
14. Egyptian used to measure the flood level in
the river __________.
15. The people of Euphrate and Tigris valleys
were aware of calendar and had the knowledge of __________.
16. The people of indus valley were pioneers of
__________ system.
17. There are two main branches of physics,
namely classical physics and __________ physics.
18. The founder of analytical algebra was
__________.
19. The author of Alsh-Shifa was __________.
20. A high precision device for measuring the
time with tremendously large accuracy is the __________.
21. __________ is the unit of thermodynamic
temperature.
22. Ampere is the unit of __________.
23. __________ is the amount of substance of a
system which contain as many elementary entities as there are atom in 0.012kg
of carbon 12.
24. The word dimension is used to denote the
__________ of a physical quantity.
25. The dimension of area is __________.
26. The dimension of linear velocity is
__________.
27. The dimension of acceleration is __________.
28. The dimension of torque is __________.
29. Dimension of universal gravitational
constant (G) is __________.
30. A number 4.71 contained __________
significant figures.
31. Some concepts of static electricity were
introduced by __________.
32. Electromagnetic wave theory was proposed by
__________.
33. The name of Pakistani physicist renowned all
over the world for his nuclear research is __________.
34. The dimension of force is __________.
35. Kelvin the unit of thermodynamic temperature
is 1/273.16 of the thermodynamic temperature of __________ freezing point of
water.
Chapter
2 - Scalars and Vectors
1. A physical quantity, which can be completely
specified by its magnitude only, is called __________.
2. A physical quantity, which can be completely
described by its magnitude and direction, is called __________.
3. Displacement is a __________ quantity.
4. Power is a __________ quantity.
5. Two vectors are __________ when they have
same magnitude and same direction.
6. The magnitude of vector will always be
__________.
7. In parallelogram law of vector addition the
resultant of two vectors is represented by __________ of the parallelogram.
8. Normally law of cosine is used to determine
the __________ of the resultant vector.
9. Law of __________ is normally used to
determine the direction of resultant vector.
10. The magnitude of __________ will always be
equal to unity.
11. The rectangular unit vectors are mutually
__________ to each other.
12. The dot product of two vectors is a
__________.
13. The __________ product of two vectors is a
vector.
14. The scalar product can be defined as that it
is the product of magnitude of two vectors and __________ of the angle between
them.
15. The vector product can be defined as that it
is the product of the magnitude of two vectors and __________ of the angle
between them.
16. The dot product will not obey the __________
law for vector multiplication.
17. The cross product will not obey the
__________ law for the vector multiplication.
18. A vector, which can be displaced parallel to
itself and applied at any point, is called __________ vector.
19. Null vector can be obtained by __________ a
vector with its negative vector.
20. Zero vectors have __________ particular
direction.
21. If two vectors are __________ to each other
then their dot product is zero.
22. If two vectors are parallel to each other
then their __________ product is zero.
23. If , then the magnitude of either of the two
vector is __________.
24. The direction of resultant vector in a
vector product can be determined by the __________ rule.
25. The magnitude of the resultant of two
vectors can be __________ than the sum of the magnitudes of individual vectors.
26. The magnitude of __________ vector will
always be zero.
27. __________.
28. __________.
29. Torque is the __________ product of force
and force arm.
30. __________ is the dot product of force and
velocity.
31. A vector, in any given direction whose
magnitude is one is called __________.
32. The set of unit vectors along x, y and z
axes are called __________ unit vectors.
33. If the magnitude of the resultant of two
vectors equal in magnitude is the same, then the angle between the two vectors
is __________.
34. If __________.
35. Two forces, one of 6N and the other of 8N,
act on a point at angle 90° with each other, the magnitude of resultant force
is __________.
Chapter 3 - Motion
1. The change of position of a body in a
particular direction is called its __________.
2. The change of displacement of a body with
respect to time is called its __________.
3. The rate of change of position in a
particular direction is called the __________ of the body.
4. The total change in displacement divided by
the total change in time of a body is called its __________.
5. The acceleration of a body is uniform when
average and instantaneous values of the acceleration are __________.
6. If the body moves towards earth, neglecting
air resistance and small changes in acceleration with altitude. This body is
referred to as free falling body and its motion is called __________.
7. __________ is an agent which changes or tends
to change the state of the body.
8. __________ is the property of the matter due
to which it will try to remain in the same state.
9. When a constant force is applied on a body
then it will start to move with constant __________.
10. Mass of the body is the measure of
__________.
11. Tension in string is a __________which will
be produced in it whenever it is subjected to pull.
12. The quantity of motion produced in a body is
called its __________.
13. The momentum of a body is defined as the
product of mass and __________ of the body.
14. A system in which the particles of the
system may exert some force one upon the other because of their collision but
no external force can be applied on them is called __________ system of
interacting bodies.
15. The collision between the two bodies is the
__________ if total momentum as well as the kinetic energy of the system remain
the same.
16. In inelastic collision the kinetic energy of
the system will __________ conserved.
17. If a massive body will collide elastically
with a lighter body at rest then the lighter body will start to move with a
velocity equal to __________ first body.
18. The force, which opposes the motion of body
on a surface, is called __________.
19. The friction is due to the __________ of the
material of the surfaces in contract.
20. The force of friction always acts __________
to the surfaces in contact.
21. The direction of the friction will always be
__________ to the direction of motion.
22. The friction within a fluid is called its
__________.
23. The friction is said to be __________ when
the body slides over the other.
24. Sliding friction is __________ than the
rolling friction.
25. The ratio of limiting friction to the normal
reaction acting between two surfaces in contact is called __________.
26. In bicycles the sliding friction is replaced
by __________.
27. On frictionless inclined plane the
acceleration of the body is __________ of its mass.
28. On frictionless inclined plane where a =
gsinq if q = 90° then a = __________.
29. On an inclined plane the ‘wcosq’ will always
be equal to __________.
30. According to the second law of motion
average force applied on the body is equal to its rate of change of __________.
31. __________ friction arises when one solid
object is set into motion across the surface of another solid body.
32. __________ friction results when a solid
object moves through fluid.
33. 1 newton = __________ dynes.
34. The acceleration produced in the body if the
applied force is equal to the weight of the body is __________.
35. The law of conservation of momentum is
equivalent to the Newton’s __________ law of motion.
36. The motion of jet plane and rocket are based
on conservation of __________.
37. When the friction between the surface of an
inclined plane and a body on it is equal to the component of the weight of the
body __________ to the plane, then the body remains at rest on the inclined
plane.
38. The acceleration of body on a smooth
inclined plane is maximum when the plane is making an angle of __________ with
the horizontal.
39. The terminal velocity of a spherical body in
fluid is inversely proportional to the __________ of the body.
40. In fluid the acceleration of spherical body
becomes zero when the force of gravity is equal to the __________ force offered
by the fluid
Chapter 4 - Motion in Two Dimensions
1. The motion of the object along a straight
line is called __________.
2. The motion of the object along a curved path
is called __________.
3. Projectile motion is the example of the
motion of the body in __________ dimension.
4. In projectile motion the object is purely under
the influence of __________.
5. In projectile motion the acceleration due to
gravity (g) will always be taken as __________.
6. During the projectile motion the horizontal
component of its velocity will __________.
7. During the projectile motion the vertical
component of its velocity is always __________.
8. The path of the projectile is a __________
path.
9. During projectile motion the acceleration
along horizontal direction is __________.
10. In projectile the acceleration along
vertical is always __________
11. In projectile the acceleration along
vertical direction is equal to the __________.
12. The expression for the time to reach the
maximum height of the projectile is __________.
13. The expression for the __________ is
2Vosinq/g.
14. The expression for the horizontal range of
the projectile is __________.
15. The expression for the maximum range of the
projectile is __________.
16. For the maximum horizontal range of the
projectile the angle of elevation must be __________.
17. The horizontal range of the projectile is
directly proportional to the square of __________.
18. The horizontal range of the projectile is
directly proportional to the sine of the twice of __________.
19. The expression for the __________ of the
projectile is ax – ½ bx2.
20. In projectile the small angle of elevation
produces __________ trajectory.
21. In projectile the __________ angle of the
elevation produces the high trajectory.
22. If the angle of elevation of the projectile
is 90° then its horizontal range is __________.
23. For the projectile with __________
trajectory their time of flight will be short.
24. If projectile has some range at an angle of
elevation of 15° then it range will be same when the angle of elevation is __________.
25. At maximum height the vertical component of
the velocity of the projectile is __________.
26. Horizontal motion with constant velocity and
vertical motion with constant acceleration is called __________.
27. If a particle is moving with constant speed
along a circle then its motion is uniform __________.
28. The angle subtended by a particle with a
centre of the circle when it is moving from one point to another on its
circumference is called angular __________.
29. The unit of angular shift is __________.
30. The angular shift per unit time of the
particle is called its angular __________.
31. The direction of angular velocity can be
determined by __________ rule.
32. The unit of angular velocity is __________.
33. If a particle covers equal angular
displacement in equal intervals of time then its angular velocity is
__________.
34. The rate of change of __________ is called
angular acceleration.
35. The velocity of the particle tangent to its
circular path is called __________.
36. In circular motion the time period and
angular velocity of a particle are __________ proportional to each other.
37. If a particle is moving with constant speed
along the circumference of a circle then the acceleration produced in it is
called __________.
38. The direction of centripetal acceleration
will always be towards the __________ of the circle.
39. Centripetal force is also called __________
force.
40. The tangential component of acceleration
arises when the speed of the particle is __________.
41. In circular motion the centripetal component
of acceleration arises when the __________ is changed.
42. Centripetal and tangential acceleration are
always __________ to each other.
43. When an object moves around the circular
track, the centripetal force is provided by __________ force.
Chapter 5 - Torque, Angular Momentum and Equilibrium
1. A body is said to be in equilibrium if it is
at rest or is moving with __________ velocity.
2. If a body is moving with uniform velocity
then the body is said to be in __________ equilibrium.
3. For the transnational equilibrium the net
force acting on the body must be __________.
4. For the rotational equilibrium, the net
torque acting on the body must be __________.
5. The body is said to be in complete
equilibrium if __________ acceleration as well as angular __________ is zero.
6. According to the first condition of
equilibrium the algebraic sum of all the forces acting on a body must be equal
to __________.
7. If the algebraic sum of all torque acting on
a body is equal to zero then the body will be in __________ equilibrium.
8. The magnitude of torque is equal to the
product of magnitude of force and its __________.
9. The unit of torque is SI system is
__________.
10. The vector product of force and displacement
is known as __________.
11. The clockwise torque is taken as __________
torque.
12. If the lines of action of the two forces
acting on the body are not same then the body will be in __________
equilibrium.
13. The physical quantity, which tends to rotate
a body, is called __________.
14. The angular momentum of a body is conserved
if the net __________ on it is zero.
15. The force which cannot be replaced by a
single equivalent force are said to form a __________.
16. __________ is defined as the time rate of
change of angular momentum.
17. In the system international the units of
angular momentum are __________.
18. The angular momentum of an isolated system
is __________.
19. A body is said to be in __________
equilibrium, if the net force on it is zero.
20. If the net torque on a body is zero then the
body is said to be __________ equilibrium.
21. A __________ body is that in which different
particles always maintain the same position relative to each other, whether the
body is at rest or in motion.
22. In rotational motion, different particles of
the body perform circular motion and the centers of all these circular orbits
lie along a straight line called the __________.
23. If a rigid body rotates with a constant
angular velocity w, then different particles of the body perform uniform
circular motion with the __________ angular velocity but __________ linear
velocities.
24. The moment of inertia of a rigid body about
a given axis of rotation is equal to the sum of the products of the mass of
each particle of the body and the square of its __________ from the given axis.
25. The __________ of a rigid body about a given
axis of rotation is given by I = Smr2 where ‘m’ is the mass of any particle of
the body situated at a distance ‘r’ from the axis.
26. The angular momentum of a particle is
defined as the __________ product of the position vector and the linear
momentum of the particle.
27. The angular momentum L of a particle is
given in terms of m, v, r and q as L = __________.
28. The direction of the angular momentum of a
particle lies along the __________ to the plane formed by the vectors and .
29. The angular momentum of a particle can be
expressed in the determinant form as = __________.
30. The direction of the angular momentum of a
particle moving with velocity v in a circular orbit of radius r is __________
to the direction of the angular velocity w which lies along the axis of
rotation.
31. The centre of gravity of uniform circular
hoop is at the __________ of the hoop.
32. The angular momentum is associated with
__________ motion.
33. The total angular momentum is associated
with __________ motion.
34. The total angular momentum of a system of
particles is __________ if the net external torque acting on the system is
zero..
35. The time rate of charge of angular momentum
of a body is equal to the __________.
Chapter 6 - Gravitation
1. According to the law of gravitation, every
body in the universe attracts every other body with a force that is directly proportional
to the __________ and inversely proportional to the square of the distance
between their centres.
2. If the distance between two bodies is
doubled, the gravitational force between them decreases to __________ of its
former value.
3. If the distance between two bodies is halved,
the force of attraction between them increases to __________ its former value.
4. The acceleration due to gravity is inversely
proportional to the __________ of the distance from centre of earth.
5. The force of attraction, which the earth
exerts on a body, is called __________.
6. Spring balance is used to determine
__________ of a body.
7. The point of body at which whole weight of
the body seems to act is called __________.
8. If the force acting on a body is equal to its
weight then the acceleration produced in the body is equal to __________.
9. The centre of gravity of a body of
rectangular shape is the point where its __________ intersect each other.
10. __________ is the centre of solar system.
11. The distance from the centre of the earth to
the centre of the moon is called __________.
12. The time taken by the earth to complete one
revolution around the sun is __________.
13. The direction of weight is always toward the
centre of the __________.
14. Newton made the hypothesis that every body
in the universe __________ the other body.
15. The moon is freely falling to __________.
16. The time taken by the moon to complete one
revolution around the earth is __________.
17. The acceleration of the moon is __________.
18. The moon’s orbit is about __________.
19. Acceleration due to gravity decreases at
__________ rate for a point above the surface of earth then for the same point
below the surface of the earth.
20. Weight of a body is a __________ quantity.
21. The __________ is responsible for the motion
of the planets around the sun.
22. Artificial gravity is produced in a
satellite by spinning its own __________.
23. The value of ‘g’ is inversely proportional
to the __________ of the radius of the earth.
24. If the mass of earth becomes four times then
the value of ‘g’ will be __________.
25. The acceleration due to gravity on moon is
__________ the acceleration due to gravity on the surface of the earth.
26. Acceleration due to gravity at the centre of
the earth is __________.
27. The weight of a body at the pole is
__________ than at equator.
28. The gravitational pull at equator is
__________ than at poles.
29. The force required to prevent a body from
falling or accelerating in a frame of reference is called the __________ of the
body.
30. An astronaut in a satellite orbiting around
the earth experiences a state of __________ because the satellite orbiting with
a centripetal acceleration equal to ‘g’.
31. Artificial gravity is produced in the
satellite to overcome the __________ in the satellite.
32. Artificial gravity is produced in the
satellite by __________ it about its own axis.
33. The value of the average density of the
earth is __________ kg/m3.
34. The dimension of gravitation constant is
__________.
35. The expression for the frequency of rotation
of the satellite to produce artificial gravity is __________.
Chapter 7
- Work, Energy and Power
1. The nature of work is positive when the force and
displacement are in the __________ direction.
2. The dot product of force and velocity is
called __________.
3. Work is __________ product of force and
displacement.
4. Power is a __________ quantity.
5. Work is a __________ quantity.
6. When the force and displacement are in the
same direction then the work is __________.
7. When the force and displacement are in the
same direction then the work is __________.
8. The ability of a body to perform the work is
called its __________.
9. The rate of doing work of a body is called its
__________.
10. Work energy equation is simply the law of
conservation of __________.
11. One joule is equal to __________ erg.
12. The unit of power is __________.
13. The unit of energy of elementary particles
and atoms is __________.
14. The energy possessed by a body by the virtue
of its motion is called the __________ energy.
15. The potential energy of a body is due to its
change of __________.
16. The work done against the gravitation field
is always __________.
17. The law of conservation of energy states
that __________ can neither be created nor destroyed but it can changed its
form.
18. The rate of change of momentum multiplied by
displacement is called __________.
19. Work done in the gravitation field is
__________ of the path followed.
20. Work done in the gravitation field along a
closed path is equal to __________.
21. The source of tidal energy is the rotation
of the earth around the __________.
22. The source of geothermal energy is the
__________ in the earth’s interior.
23. The source of nuclear energy is __________
in heavy nucleus.
24. Einstein mass energy equation is E =
__________.
25. The unit of energy is the same as the unit
of __________.
26. The dimensions of work are __________.
27. ML2T-3 are the dimensions of __________.
28. The rate of expenditure of energy is called
__________.
29. The amount of work done in lifting a body
from the surface of the earth to the zero point is called __________ energy.
30. One horsepower is equal to __________ watt.
31. One horsepower is equal to __________
ft.lb/s
32. The expression for absolute gravitational
potential energy is __________.
33. 1 electron volt = joule.
34. 1 giga watt (GW) = __________ W.
35. 1 Kilowatt hour (1kWh) = __________ joules
Chapter 8
- Wave Motion and Sound
1. The SHM is a __________ motion.
2. The SHM is a motion under __________ force.
3. In case of SHM __________ is proportional to
displacement and is always directed towards mean position.
4. In SHM the system should be free from
__________.
5. In SHM the system must possess __________.
6. In SHM the amplitude of vibration remains
__________.
7. In SHM the time period of vibrating body
remain __________.
8. The time period of simple pendulum __________
with its length.
9. The gravitational acceleration at a place is
inversely proportional to the __________ of its time period of a simple
pendulum at that place.
10. The simple pendulum comes to rest after some
time due to the friction at the support and __________ resistance.
11. Stationary waves are produced due to the
phenomenon of __________.
12. In stationary waves, node are the points at
which __________ of the particles of the medium is zero.
13. In standing waves, the points at which
particles of medium vibrate maximum amplitude are called __________.
14. In standing waves, at nodes particles, of
the medium are under __________ tension.
15. In standing waves produced in a stretched
string, when string is vibrating in two loops, the frequency of waves is equal
to the __________ of fundamental frequency.
16. In standing waves produced in a stretched
string, the second harmonic is also called __________ multiple.
17. Decibel is a unit of __________.
18. The number of beats produced per second is
equal to __________ of frequency between the two vibrating bodies producing
beats.
19. The phenomenon in the variation of
__________ of sound due to relative motion of source or observer is called
Doppler’s Effect.
20. When a listener moves towards a stationary
source, the pitch of sound appears to __________.
21. The equation Y = f(x-vt) represents a
travelling wave travelling towards __________.
22. The sound waves are __________ and
longitudinal waves in nature.
23. The speed of sound waves through a gas is
__________ to its absolute temperature.
24. The particle’s velocity at antinodes in
standing wave is __________.
25. The pitche of a sound note depends upon its
__________ of vibration.
26. Y = -Aosin kx (coswt) represents the
equation of a __________ wave.
27. Another term used for quality of sound is
__________.
28. The characteristic by which one sound can be
distinguished from other is known as __________.
29. The sound absorbing soft porous materials
are used in big hall is order to have good __________.
30. Radar is an abbreviation of __________.
31. In standing waves the distance between two
__________ nodes or antinodes is equal to one wavelength.
32. In standing waves the distance between one
node and nearest antinode is equal to __________ wavelength.
33. In standing waves the particles velocity at
the nodal point is __________.
34. The velocity of sound waves in vacuum is
__________.
35. When sound waves travel through any gas the
phenomena is __________.
Chapter 9 - Wave Aspect of Light
1. Newton supported __________ theory of light.
2. __________ suggested wave theory of light.
3. __________ formulated electromagnetic theory.
4. Photon is a __________ of light.
5. Wave theory predicts __________ velocity of
light in material medium then vacuum.
6. The first clear demonstration that light is a
wave phenomenon was made by __________.
7. The electromagnetic radiation having
wavelength between 4000A° and 7000A° are called __________.
8. The electromagnetic radiation of wavelength
more than 7000A° are known as __________.
9. Photoelectric effect and Compton effect
supported the __________ theory of light.
10. Interference of light can be explained on
the basis of __________ theory of light.
11. The locus of all points, which are in same
state of vibration, is called __________.
12. The points of constructive interference are
always of __________ intensity.
13. The points of destructive interference are
of __________ intensity.
14. A small portion of a spherical wave front,
at a large distance from the source becomes almost.
15. The shape of wave front at a very small
distance from the source of light is __________.
16. For points of constructive interference the
path difference between two interfering wave is zero of __________ of the
wavelength.
17. For the points of __________ interference
the path difference between two interfering waves is odd multiple of half
wavelength.
18. If we decrease the distance between slits
and screen, in Young’s Experiment, fringe spacing will __________.
19. If we decrease the separation between the
slits in Young’s Experiment the fringe will __________.
20. In Young’s Experiment the fringes are
equally __________.
21. In Young’s experiment the __________ fringe
is always bright.
22. A glass plate on which equally spaced lines
are ruled is known as __________.
23. The distance between two consecutive slits
of diffraction during grating is known as __________.
24. To obtain Newton’s Rings a __________ lens
of __________ focal length is required.
25. Newton’s rings are formed due to the phenomena
of __________ interference.
26. In __________ the central ring may be bright
or dark.
27. In thin film interference for destructive
interference between two interfering waves be __________ or __________ of
wavelength.
28. In Young’s Experiment two narrow sites are
used to obtain __________.
29. In Michlson’s interferometer the purpose of
beam splitter is to obtain __________.
30. In thin film interference phase coherence is
obtained by reflecting the light from __________ and __________ surface of the
air film.
31. Newton’s rings are formed due to __________
film formed between convex surface and plane glass sheet.
32. The speed of X-rays is __________ to the
speed of light.
33. __________ is a special type of
interference.
34. The phenomenon of polarization confirms the
__________ of light.
35. X-ray diffraction can be observed when they
are made incident on __________.
36. In case of Young’s Experiment the distance
between two consecutive bright or dark fringes is known as __________.
37. The bending of light around an obstacle or
sharp edge is called __________.
38. The phenomenon of polarization can be used
to check __________ of optically active substances in a solution.
39. The points at which two sources cancel each
other effects are known as points __________ interference.
40. If l is the wavelength of light in a medium
of refractive index n then wavelength of light in air l is equal to __________
Work, Power and Energy
Work
Work is said to be done when a force causes a
displacement to a body on which it acts.
Work is a scalar quantity. It is the dot product
of applied force F and displacement d.
Diagram Coming Soon
W = F . d
W = F d cos θ .............................. (1)
Where θ is the angle between F and d.
Equation (1) can be written as
W = (F cos θ) d
i.e., work done is the product of the component
of force (F cos θ) in the direction of displacement and the magnitude of
displacement d.
equation (1) can also be written as
W = F (d cos θ)
i.e., work done is the product of magnitude of
force F and the component of the displacement (d cos θ) in the direction of
force.
Unit of Work
M.K.S system → Joule, BTU, eV
C.G.S system → Erg
F.P.S system → Foot Pound
1 BTU = 1055 joule
1 eV = 1.60 x 10(-19)
Important Cases
Work can be positive or negative depending upon
the angle θ between F and d.
Case I
When θ = 0º i.e., when F and d have same
direction.
W = F . d
W = F d cos 0º .............. {since θ = 0º}
W = F d .......................... {since cos 0º
= 1}
Work is positive in this case.
Case II
When θ = 180º i.e., when F and d have opposite
direction.
W = F . d
W = F d cos 180º ............................
{since θ = 180º}
W = - F d .........................................
{since cos 180º = -1}
Work is negative in this case
Case III
When θ = 90º i.e, when F and d are mutually
perpendicular.
W = F . d
W = F d cos 90º ............................
{since θ = 90º}
W = 0 ...........................................
{since cos 90º = 0}
Work Done Against Gravitational Force
Consider a body of mass 'm' placed initially at
a height h(i), from the surface of the earth. We displaces it to a height h(f)
from the surface of the earth. Here work is done on the body of mass 'm' by
displacing it to a height 'h' against the gravitational force.
W = F . d = F d cos θ
Here,
F = W = m g
d = h(r) - h(i) = h
θ = 180º
{since mg and h are in opposite direction}
Since,
W = m g h cos 180º
W = m g h (-1)
W = - m g h
Since this work is done against gravitational
force, therefore, it is stored in the body as its potential energy (F.E)
Therefore,
P . E = m g h
Power
Power is defined as the rate of doing work.
If work ΔW is done in time Δt by a body, then
its average power is given by P(av) = ΔW / Δt
Power of an agency at a certain instant is
called instantaneous power.
Relation Between Power and Velocity
Suppose a constant force F moves a body through
a displacement Δd in time Δt, then
P = ΔW / Δt
P = F . Δd / Δt ..................... {since ΔW
= F.Δd}
P = F . Δd / Δt
P = F . V .................................
{since Δd / Δt = V}
Thus power is the dot product of force and
velocity.
Units of Power
The unit of power in S.I system is watt.
P = ΔW / Δt = joule / sec = watt
1 watt is defined as the power of an agency
which does work at the rate of 1 joule per second.
Bigger Units → Mwatt = 10(6) watt
Gwatt = 10(9) watt
Kilowatt = 10(3) watt
In B.E.S system, the unit of power is
horse-power (hp).
1 hp = 550 ft-lb/sec = 746 watt
Energy
The ability of a body to perform work is called
its energy. The unit of energy in S.I system is joule.
Kinetic Energy
The energy possessed by a body by virtue of its
motion is called it kinetic energy.
K.E = 1/2 mv2
m = mass,
v = velocity
Prove K.E = 1/2 mv2
Proof
Kinetic energy of a moving body is measured by
the amount of work that a moving body can do against an unbalanced force before
coming to rest.
Consider a body of mas 'm' thrown upward in the
gravitational field with velocity v. It comes to rest after attaining height
'h'. We are interested in finding 'h'.
Therefore, we use
2 a S = vf2 - vi2 ............................
(1)
Here a = -g
S = h = ?
vi = v (magnitude of v)
vf = 0
Therefore,
(1) => 2(-g) = (0)2 - (v)2
-2 g h = -v2
2 g h = v2
h = v2/2g
Therefore, Work done by the body due to its
motion = F . d
= F d cos θ
Here
F = m g
d = h = v2 / 2g
θ = 0º
Therefore, Work done by the body due to its
motion = (mg) (v2/2g) cos0º
= mg x v2 / 2g
= 1/2 m v2
And we know that this work done by the body due
to its motion.
Therefore,
K.E = 1/2 m v2
Potential Energy
When a body is moved against a field of force,
the energy stored in it is called its potential energy.
If a body of mass 'm' is lifted to a height 'h'
by applying a force equal to its weight then its potential energy is given by
P.E = m g h
Potential energy is possessed by
1. A spring when it is compressed
2. A charge when it is moved against
electrostatic force.
Prove P.E = m g h OR Ug = m g h
Proof
Consider a ball of mass 'm' taken very slowly to
the height 'h'. Therefore, work done by external force is
Wex = Fex . S = Fex S cos θ
.................................. (1)
Since ball is lifted very slowly, therefore,
external force in this case must be equal to the weight of the body i.e., mg.
Therefore,
Fex = m g
S = h
θ = 0º ......................... {since Fex and
h have same direction}
Therefore,
(1) => Wex = m g h cos 0º
Wex = m g h ..................................................
................. (2)
Work done by the gravitational force is
Wg = Fg . S = Fg S cosθ
................................................. (3)
Since,
Fg = m g
S = h
θ = 180º ...................... {since Fg and h
have opposite direction}
Therefore,
(3) => Wg = m g h cos 180º
Wg = m g h (-1)
Wg = - m g h
.................................................. ................. (4)
Comparing (2) and (4), we get
Wg = -Wex
Or
Wex = - Wg
The work done on a body by an external force
against the gravitational force is stored in it as its gravitational potential
energy (Ug).
Therefore,
Ug = Wex
Ug = - Wg .............................. {since
Wex = -Wg}
Ug = -(-m g h) ..................... {since Wg =
- m g h}
Ug = m g h ...............................................
Proved
Absolute Potential Energy
In gravitational field, absolute potential
energy of a body at a point is defined as the amount of work done in moving it
from that point to a point where the gravitational field is zero.
Determination of Absolute Potential Energy
Consider a body of mass 'm' which is lifted from
point 1 to point N in the gravitational field. The distance between 1 and N is
so large that the value of g is not constant between the two points. Hence to
calculate the work done against the force of gravity, the simple formula W = F
.d cannot be applied.
Therefore, in order to calculate work done from
1 to N, we divide the entire displacement into a large number of small
displacement intervals of equal length Δr. The interval Δr is taken so small
that the value of g remains constant during this interval.
Diagram Coming Soon Now we calculate the work
done in moving the body from point 1 to point 2. For this work the value of
constant force F may be taken as the average of the forces acting at the ends
of interval Δr. At point 1 force is F1 and at point 2, force is F2.
Law of Conservation of Energy
Statement
Energy can neither be created nor be destroyed, however, it can be
transformed from one form to another.
Explanation
According to this law energy may change its form
within the system but we cannot get one form of energy without spending some
other form of energy. A loss in one form of energy is accompanied by an
increase in other forms of energy. The total energy remains constant.
Proof
For the verification of this law in case of
mechanical energy (Kinetic and potential energy). Let us consider a body of
mass 'm' placed at a point P which is at a height 'h' from the surface of the
earth. We find total energy at point P, point O and point Q. Point Q is at a
distance of (h-x) from the surface of earth.
Gravitation
Gravitation
The property of all objects in the universe
which carry mass, by virtue of which they attract one another, is called
Gravitation.
Centripetal Acceleration of the Moon
Newton, after determining the centripetal
acceleration of the moon, formulated the law of universal gravitation.
Suppose that the moon is orbiting around the
earth in a circular orbit.
If V = velocity of the moon in its orbit,
Rm = distance between the centres of earth and
moon,
T = time taken by the moon to complete one
revolution around the earth.
For determining the centripetal acceleration of
the moon,. Newton applied Huygen's formula which is
a(c) = v2 / r
For moon, am = v2 / Rm ..................... (1)
But v = s/t = circumference / time period =
2πRm/T
Therefore,
v2 = 4π2Rm2 / T2
Therefore,
=> a(m) = (4π2Rm2/T2) x (1/Rm)
a(m) = 4π2Rm / T2
Put Rm = 3.84 x 10(8) m
T = 2.36 x 10(6) sec
Therefore,
a(m) = 2.72 x 10(-3) m/s2
Comparison Between 'am' AND 'g'
Newton compared the centripetal acceleration of
the moon 'am' with the gravitational acceleration 'g'.
i.e., am / g = 1 / (60)2 ................. (1)
If Re = radius of the earth, he found that
Re2 / Rm2 - 1 / (60)2 .........................
(2)
Comparing (1) and (2),
am / g = Re2 / Rm2
..................................... (3)
From equation (3), Newton concluded that at any
point gravitational acceleration is inversely to the square of the distance of
that point from the centre of the earth. It is true of all bodies in the
universe. This conclusion provided the basis for the Newton' Law of Universal
Gravitation.
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Newton's Law of Universal Gravitation
Consider tow bodies A and B having masses mA and
mB respectively.
Diagram Coming Soon
Let,
F(AB) = Force on A by B
F(BA) = Force on B by A
r(AB) = displacement from A to B
r(BA) = displacement from B to A
r(AB) = unit vector in the direction of r(AB).
r(BA) = unit vector in the direction of r(BA).
From a(m) / g = Re2 / Rm2, we have
F(AB) ∞ 1 / r(BA)2 ......................... (1)
Also,
F(AB) ∞ m(A) ............................... (2)
F(BA) ∞ m(B)
According to the Newton's third law of motion
F(AB) = F(BA) .................... (for
magnitudes)
Therefore,
F(AB) ∞ m(B) ................................
(3)
Combining (1), (2) and (3), we get
F(AB) ∞ m(A)m(B) / r(BA)2
F(AB) = G m(A)m(B) / r(BA)2 ...........................
(G = 6.67 x 10(-11) N - m2 / kg2)
Vector Form
F(AB) = - (G m(A)m(B) / r(BA)2) r(BA)
F(BA) = - (G m(B)m(A) / r(AB)22) r(AB)
Negative sign indicates that gravitational force
is attractive.
Statement of the Law
"Every body in the universe attracts every
other body with a force which is directly proportional to the products of their
masses and inversely proportional to the square of the distance between their
centres."
Mass and Average Density of Earth
Let,
M = Mass of an object placed near the surface of
earth
M(e) = Mass of earth
R(e) = Radius of earth
G = Acceleration due to gravity
According to the Newton's Law of Universal
Gravitation.
F = G M Me / Re2 .............................
(1)
But the force with which earth attracts a body
towards its centre is called weight of that body.
Diagram Coming Soon
Therefore,
F = W = Mg
(1) => M g = G M Me / Re2
g = G Me / Re2
Me = g Re2 / G
.................................. (2)
Put
g = 9.8 m/sec2,
Re = 6.38 x 10(6) m,
G = 6.67 x 10(-11) N-m2/kg2, in equation (2)
(1) => Me = 5.98 x 10(24) kg.
................... (In S.I system)
Me = 5.98 x 10(27) gm
.................................... (In C.G.S system)
Me = 6.6 x 10(21) tons
For determining the average density of earth
(ρ),
Let Ve be the volume of the earth.
We know that
Density = mass / volume
Therefore,
ρ = Me / Ve ........................... [Ve =
volume of earth]
ρ = Me / (4/3 Π Re3) .............. [since Ve =
4/3 Π Re3]
ρ = 3 Me / 4 Π Re3
Put,
Me = 5.98 x 10(24) kg
& Re = 6.38 x 10(6) m
Therefore,
ρ = 5.52 x 10(3) kg / m3
Mass of Sun
Let earth is orbiting round the sun in a
circular orbit with velocity V.
Me = Mass of earth
Ms = Mass of the sun
R = Distance between the centres of the sun and
the earth
T = Period of revolution of earth around sun
G = Gravitational constant
According to the Law of Universal Gravitation
F = G Ms Me / R2
.................................... (1)
This force 'F' provides the earth the necessary
centripetal force
F = Me V2 / R
............................................ (2)
(1) & (2) => Me V2 / R = G Ms Me / R2
=> Ms = V2 R / G
........................................ (3)
V = s / Π = 2Π R / T
=> V2 = 4Π2 R2 / T2
Therefore,
(3) => Ms = (4Π R2 / T2) x (R / G)
Ms = 4Π2 R3 / GT2
........................................ (4)
Substituting the value of
R = 1.49 x 10(11),
G = 6.67 x 10(-11) N-m2 / kg2,
T = 365.3 x 24 x 60 x 60 seconds, in equation
(4)
We get
Ms = 1.99 x 10(30) kg
Variation of 'g' with Altitude
Suppose earth is perfectly spherical in shape
with uniform density ρ. We know that at the surface of earth
g = G Me / Re2
where
G = Gravitational constant
Me = Mass of earth
Re = Radius of earth
At a height 'h' above the surface of earth,
gravitational acceleration is
g = G Me / (Re + h)2
Dividing (1) by (2)
g / g = [G Me / Re2] x [(Re + h)2 / G Me)
g / g = (Re + h)2 / Re2
g / g = [Re + h) / Re]2
g / g = [1 + h/Re]2
g / g = [1 + h/Re]-2
We expand R.H.S using Binomial Formula,
(1 + x)n = 1 + nx + n(n-1) x2 / 1.2 + n(n +
1)(n-2)x3 / 1.2.3 + ...
If h / Re < 1, then we can neglect higher
powers of h / Re.
Therefore
g / g = 1 - 2 h / Re
g = g (1 - 2h / Re)
................................. (3)
Equation (3) gives the value of acceleration due
to gravity at a height 'h' above the surface of earth.
From (3), we can conclude that as the value of
'h' increases, the value of 'g' decreases.
Variation of 'g' with Depth
Suppose earth is perfectly spherical in shape
with uniform density ρ.
Let
Re = Radius of earth
Me = Mass of earth
d = Depth (between P and Q)
Me = Mass of earth at a depth 'd'
At the surface of earth,
g = G Me / Re2
...................................... (1)
At a depth 'd', acceleration due to gravity is
g = G Me / (Re - d)2 ...........................
(2)
Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ
Me = ρ x Ve = ρ x (4/3) π (Re - d)3 = 4/3 π (Re
- d)3 ρ
Ve = Volume of earth
Substitute the value of Me in (1),
(1) => g = (G / Re2) x (4/3) π Re3 ρ
g = 4/3 π Re ρ G
................................. (3)
Substitute the value of Me in (2)
g = [G / Re - d)2] x (4/3) π (Re - d)3 ρ
g = 4/3 π (Re - d) ρ G
Dividing (4) by (3)
g / g = [4/3 π (Re - d) ρ G] / [4/3 π Re ρ G]
g / g = (Re - d) / Re
g / g = 1 - d/Re
g / g = g (1 - d / Re)
........................... (5)
Equation (5) gives the value of acceleration due
to gravity at a depth 'd' below the surface of earth
From (5), we can conclude that as the value of
'd' increases, value of 'g' decreases.
At the centre of the earth.
d = Re => Re / d = 1
Therefore,
(5) => g = g (1-1)
g = 0
Thus at the centre of the earth, the value of
gravitational acceleration is zero.
Weightlessness in Satellites
An apparent loss of weight experienced by a body
in a spacecraft in orbit is called weightlessness.
To discuss weightlessness in artificial
satellites, let us take the example of an elevator having a block of mass ;m;
suspended by a spring balance attached to the coiling of the elevator. The
tension in the thread indicates the weight of the block.
Consider following cases.
1. When Elevator is at Rest
T = m g
2. When Elevator is Ascending with an Acceleration 'a'
In this case
T > m g
Therefore, Net force = T - mg
m a = T - m g
T = m g + m a
In this case of the block appears
"heavier".
3. When Elevator is Descending with an Acceleration 'a'
In this case
m g > T
Therefore
Net force = m g - T
m a = m g - T
T = m g - m a
In this case, the body appears lighter
4. When the Elevator is Falling Freely Under the Action of Gravity
If the cable supporting the elevator breaks, the
elevator will fall down with an acceleration equal to 'g'
From (3)
T = m g - m a
But a = g
Therefore
T = m g - m g
T = 0
In this case, spring balance will read zero.
This is the state of "weightlessness".
In this case gravitation force still acts on the
block due to the reason that elevator block, spring balance and string all have
same acceleration when they fall freely, the weight of the block appears zero.
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Artificial Gravity
In artificial satellites, artificial gravity can
be created by spinning the space craft about its own axis.
Now we calculate frequency of revolution (v) of
a space craft of length 2R to produce artificial gravity in it. Its time period
be 'T' and velocity is V.
Torque, Angular, Momentum and Equilibrium
Torque or Moment of Force
Definition
If a body is capable of rotating about an axis,
then force applied properly on this body will rotate it about the axis (axis of
rotating). This turning effect of the force about the axis of rotation is
called torque.
Torque is the physical quantity which produces
angular acceleration in the body.
Explanation
Consider a body which can rotate about O (axis
of rotation). A force F acts on point P whose position vector w.r.t O is r.
Diagram Coming Soon
F is resolved into F1 and F2. θ is the angle
between F and extended line of r.
The component of F which produces rotation in
the body is F1.
The magnitude of torgue (π) is the product of
the magnitudes of r and F1.
Equation (1) shows that torque is the
cross-product of displacement r and force F.
Torque → positive if directed outward from paper
Torque → negative if directed inward from paper
The direction of torque can be found by using
Right Hand Rule and is always perpendicular to the plane containing r & F.
Thus
Clockwise torque → negative
Counter-Clockwise torque → positive
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Alternate Definition of Torque
π = r x F
|π| = r F sin θ
|π| = F x r sin θ
But r sin θ = L (momentum arm) (from figure)
Therefore,
|π| = F L
Magnitude of Torque = Magnitude of force x
Moment Arm
Note
If line of action of force passes through the
axis of rotation then this force cannot produce torque.
The unit of torque is N.m.
Couple
Two forces are said to constitute a couple if
they have
1. Same magnitudes
2. Opposite directions
3. Different lines of action
These forces cannot produces transiatory motion,
but produce rotatory motion.
Moment (Torque) of a Couple
Consider a couple composed of two forces F and
-F acting at points A and B (on a body) respectively, having position vectors
r1 & r2.
If π1 is the torque due to force F, then
π1 = r1 x F
Similarly if π2 is the torque due to force - F,
then
π2 = r2 x (-F)
The total torque due to the two forces is
π = π1 + π2
π = r1 x F + r2 x (-F)
π = r1 x F - r2 x (-F)
π = (r1 - r2) x F
π = r x F
where r is the displacement vector from B to A.
The magnitude of torque is
π = r F sin (180 - θ)
π = r F sin θ .................... {since sin
(180 - θ) = sin θ}
Where θ is the angle between r and -F.
π = F (r sin θ)
But r sin θ is the perpendicular distance
between the lines of action of forces F and -F is called moment arm of the couple denoted by d.
π = Fd
Thus
[Mag. of the moment of a couple] = [Mag. of any
of the forces forming the couple] x [Moment arm of the couple]
Moment (torque) of a given couple is independent
of the location of origin.
Centre of Mass
Definition
The centre of mass of a body, or a system of
particles, is a point on the body that moves in the same way that a single
particle would move under the influence of the same external forces. The whole
mass of the body is supposed to be concentrated at this point.
Explanation
During translational motion each point of a body
moves in the same manner i.e., different particles of the body do not change
their position w.r.t each other. Each point on the body undergoes the same
displacement as any other point as time goes on. So the motion of one particle
represents the motion of the whole body. But in rotating or vibrating bodies
different particles move in different manners except one point called centre of
mass. The centre of mass of a body or a system of particle is a point which
represents the movement of the entire system. It moves in the same way that a
single particle would move under the influence of same external forces.
Centre of Mass and Centre of Gravity
In a completely uniform gravitational field, the
centre of mass and centre of gravity of an extended body coincides. But if
gravitational field is not uniform, these points are different.
Determination of the Centre of Mass
Consider a system of particles having masses m1,
m2, m3, ................. mn. Suppose x1, and z1, z2, z3 are their distances on
z-axis, all measured from origin.
Equilibrium
body is
said to be in equilibrium if it is
1. At rest, or
2. Moving with uniform velocity
A body in equilibrium possess no acceleration.
Static Equilibrium
The equilibrium of bodies at rest is called
static equilibrium. For example,
1. A book lying on a table
2. A block hung from a string
Dynamic Equilibrium
The equilibrium of bodies moving with uniform
velocity is called dynamic equilibrium. For example,
1. The jumping of a paratrooper by a parachute
is an example of uniform motion. In this case, weight is balanced by the
reaction of the air on the parachute acts in the vertically upward direction.
2. The motion of a small steel ball through a
viscous liquid. Initially the ball has acceleration but after covering a
certain distance, its velocity becomes uniform because weight of the ball is
balanced by upward thrust and viscous force of the liquid. Therefore, ball is
in dynamic equilibrium.
Angular Momentum
Definition
The quantity of rotational motion in a body is
called its angular momentum. Thus angular momentum plays same role in rotational
motion as played by linear momentum in translational motion.
Mathematically, angular momentum is the
cross-product of position vector and the linear momentum, both measured in an
inertial frame of reference.
ρ = r x P
Explanation
Consider a mass 'm' rotating anti-clockwise in
an inertial frame of reference. At any point, let P be the linear momentum and
r be the position vector.
ρ = r x P
ρ = r P sinθ ........... (magnitude)
ρ = r m V sinθ .......... {since P = m v)
where,
V is linear speed
θ is the angle between r and P
θ = 90º in circular motion (special case)
The direction of the angular momentum can be
determined by the Righ-Hand Rule.
Also
ρ = r m (r ω) sin θ
ρ = m r2 ω sin θ
Units of Angular Momentum
The units of angular momentum in S.I system are
kgm2/s or Js.
1. ρ = r m V sin θ
= m x kg x m/s
= kg.m2/s
2. ρ = r P sin θ
= m x Ns
= (Nm) x s
= J.s
Dimensions of Angular Momentum
[ρ] = [r] [P]
= [r] [m] [V]
= L . M . L/T
= L2 M T-1
Relation Between Torque and Angular Momentum
OR
Prove that the rate of change of angular momentum is equal to the external
torque acting on the body.
Proof
We know that rate of change of linear momentum
is equal to the applied force.
F = dP / dt
Taking cross product with r on both sides, we
get
R x F = r x dP / dt
τ = r x dP / dt .............................
{since r x P = τ}
Now, according to the definition of angular
momentum
ρ = r x P
Taking derivative w.r.t time, we get
dρ / dt = d / dt (r x P)
=> dρ / dt = r x dP / dt + dr / dt x P
=> dρ / dt = τ + V x P ..................
{since dr / dt = V}
=> dρ / dt = τ + V x mV
=> dρ / dt = τ + m (V x V)
=> dρ / dt = τ + 0 ................. {since V
x V = 0}
=> dρ / dt = τ
Or, Rate of change of Angular Momentum =
External Torque ................. (Proved)
Motion in two Dimension Projectile Motion
A body moving horizontally as well as vertically
under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its
trajectory.
Examples of projectile motion are
1. Kicked or thrown balls
2. Jumping animals
3. A bomb released from a bomber plane
4. A shell of a gun.
Analysis of Projectile Motion
Let us consider a body of mass m, projected an
angle θ with the horizontal with a velocity V0. We made the following three
assumptions.
1. The value of g remains constant throughout
the motion.
2. The effect of air resistance is negligible.
3. The rotation of earth does not affect the
motion.
Horizontal Motion
Acceleration : ax = 0
Velocity : Vx = Vox
Displacement : X = Vox t
Vertical Motion
Acceleration : ay = - g
Velocity : Vy = Voy - gt
Displacement : Y = Voy t - 1/2 gt2
Initial Horizontal Velocity
Vox = Vo cos θ ...................... (1)
Initial Vertical Velocity
Voy = Vo sin θ ...................... (2)
Net force W is acting on the body in downward
vertical direction, therefore, vertical velocity continuously changes due to
the acceleration g produced by the weight W.
There is no net force acting on the projectile
in horizontal direction, therefore, its horizontal velocity remains constant
throughout the motion.
X - Component of Velocity at Time t (Vx)
Vx = Vox = Vo cos θ .................... (3)
Y - Component of Velocity at Time t (Vy)
Data for vertical motion
Vi = Voy = Vo sin θ
a = ay = - g
t = t
Vf = Vy = ?
Using Vf = Vi + at
Vy = Vo sin θ - gt .................... (4)
Range of the Projectile (R)
The total distance covered by the projectile in
horizontal direction (X-axis) is called is range
Let T be the time of flight of the projectile.
Therefore,
R = Vox x T .............. {since S = Vt}
T = 2 (time taken by the projectile to reach the
highest point)
T = 2 Vo sin θ / g
Vox = Vo cos θ
Therefore,
R = Vo cos θ x 2 Vo sin θ / g
R = Vo2 (2 sin θ cos θ) / g
R = Vo2 sin 2 θ / g .................. { since 2
sin θ cos θ = sin2 θ}
Thus the range of the projectile depends on
(a) The square of the initial velocity
(b) Sine of twice the projection angle θ.
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The Maximum Range
For a given value of Vo, range will be maximum
when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
0 ≤ sin2 θ ≤ 1
Hence maximum value of sin2 θ is 1.
Sin2 θ = 1
2θ = sin(-1) (1)
2θ = 90º
θ = 45º
Therefore,
R(max) = Vo2 / g ; at θ = 45º
Hence the projectile must be launched at an
angle of 45º with the horizontal to attain maximum range.
Projectile Trajectory
The path followed by a projectile is referred as
its trajectory.
We known that
S = Vit + 1/2 at2
For vertical motion
S = Y
a = - g
Vi = Voy = Vo sin θ
Therefore,
Y = Vo sinθ t - 1/2 g t2 .......................
(1)
Also
X = Vox t
X = Vo cosθ t ............ { since Vox = Vo
cosθ}
t = X / Vo cos θ
(1) => Y = Vo sinθ (X / Vo cos θ) - 1/2 g (X
/ Vo cos θ)2
Y = X tan θ - gX2 / 2Vo2 cos2 θ
For a given value of Vo and θ, the quantities
tanθ, cosθ, and g are constant, therefore, put
a = tan θ
b = g / Vo2 cos2θ
Therefore
Y = a X - 1/2 b X2
Which shows that trajectory is parabola.
Uniform Circular Motion
If an object moves along a circular path with
uniform speed then its motion is said to be uniform circular motion.
Recitilinear Motion
Displacement → R
Velocity → V
Acceleration → a
Circular Motion
Angular Displacement → θ
Angular Velocity → ω
Angular Acceleration → α
Angular Displacement
The angle through which a body moves, while
moving along a circular path is called its angular displacement.
The angular displacement is measured in degrees,
revolutions and most commonly in radian.
Diagram Coming Soon s = arc length
r = radius of the circular path
θ = amgular displacement
It is obvious,
s ∞ θ
s = r θ
θ = s / r = arc length / radius
Radian
It is the angle subtended at the centre of a
circle by an arc equal in length to its radius.
Therefore,
When s = r
θ = 1 radian = 57.3º
Angular Velocity
When a body is moving along a circular path,
then the angle traversed by it in a unit time is called its angular velocity.
Diagram Coming Soon Suppose a particle P is
moving anticlockwise in a circle of radius r, then its angular displacement at
P(t1) is θ1 at time t1 and at P(t2) is θ2 at time t2.
Average angular velocity = change in angular
displacement / time interval
Change in angular displacement = θ2 - θ1 = Δθ
Time interval = t2 - t1 = Δt
Therefore,
ω = Δθ / Δt
Angular velocity is usually measured in rad/sec.
Angular velocity is a vector quantity. Its direction
can be determined by using right hand rule according to which if the axis of
rotation is grasped in right hand with fingers curled in the direction of
rotation then the thumb indicates the direction of angular velocity.
Angular Acceleration
It is defined as the rate of change of angular
velocity with respect to time.
Thus, if ω1 and ω2 be the initial and final
angular velocity of a rotating body, then average angular acceleration
"αav" is defined as
αav = (ω2 - ω1) / (t2 - t1) = Δω / Δt
The units of angular acceleration are
degrees/sec2, and radian/sec2.
Instantaneous angular acceleration at any
instant for a rotating body is given by
Angular acceleration is a vector quantity. When
ω is increasing, α has same direction as ω. When ω is decreasing, α has
direction opposite to ω.
Relation Between Linear Velocity And Angular Velocity
Consider a particle P in an object in X-Y plane
rotating along a circular path of radius r about an axis through O,
perpendicular to the plane of figure as shown here (z-axis).
If the particle P rotates through an angle Δθ in
time Δt,
Then according to the definition of angular
displacement.
Δθ = Δs / r
Dividing both sides by Δt,
Δθ / Δt = (Δs / Δt) (1/r)
=> Δs / Δt = r Δθ / Δt
For a very small interval of time
Δt → 0
Alternate Method
We know that for linear motion
S = v t .............. (1)
And for angular motion
S = r θ ................. (2)
Comparing (1) & (2), we get
V t = r θ
v = r θ/t
V = r ω ........................... {since θ/t =
ω}
Relation Between Linear Acceleration And Angular Acceleration
Suppose an object rotating about a fixed axis,
changes its angular velocity by Δω in Δt. Then the change in tangential
velocity, ΔVt, at the end of this interval is
ΔVt = r Δω
Dividing both sides by Δt, we get
ΔVt / Δt = r Δω / Δt
If the time interval is very small i.e., Δt → 0 then
Alternate Method
Linear acceleration of a body is given by
a = (Vr - Vi) / t
But Vr = r ω r and Vi = r ω i
Therefore,
a = (r ω r - r ω i) / t
=> a = r (ωr- ωi) / t
a = r α ....................................
{since (ωr = ωi) / t = ω}
Time Period
When an object is rotating in a circular path,
the time taken by it to complete one revolution or cycle is called its time
period, (T).
We know that
ω = Δθ / Δt OR Δt = Δθ / ω
For one complete rotation
Δθ = 2 π
Δt = T
Therefore,
T = 2 π / ω
If ω = 2πf ........................ {since f =
frequency of revolution}
Therefore,
T = 2π / 2πf
=> T = 1 / f
Tangential Velocity
When a body is moving along a circle or circular
path, the velocity of the body along the tangent of the circle is called its
tangential velocity.
Vt = r ω
Tangential velocity is not same for every point
on the circular path.
Centripetal Acceleration
A body moving along a circular path changes its
direction at every instant. Due to this change, the velocity of the body 'V' is
changing at every instant. Thus body has an acceleration which is called its
centripetal acceleration. It is denoted by a(c) or a1 and always directed
towards the centre of the circle. The magnitude of the centripetal acceleration
a(c) is given as follows
a(c) = V2 / r, ........................... r =
radius of the circular path
Prove That a(c) = V2 / r
Proof
Consider a body moving along a circular path of
radius of r with a constant speed V. Suppose the body moves from a point P to a
point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is
V2. Let the angular displacement made in this time be ΔO .
Since V1 and V2 are perpendicular to the radial
lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles
OPQ and ABC are similar.
Therefore,
|ΔV| / |V1| = Δs / r
Since the body is moving with constant speed
Therefore,
|V1| = |V2| = V
Therefore,
ΔV / V = Vs / r
ΔV = (V / r) Δs
Dividing both sides by Δt
Therefore,
ΔV / Δt = (V/r) (V/r) (Δs / Δt)
taking limit Δt → 0.
Proof That a(c) = 4π2r / T2
Proof
We know that
a(c) = V2 / r
But V = r ω
Therefore,
a(c) = r2 ω2 / r
a(c) = r ω2 ...................... (1)
But ω = Δθ / Δt
For one complete rotation Δθ = 2π, Δt = T (Time
Period)
Therefore,
ω = 2π / T
(1) => a(c) = r (2π / T)2
a(c) = 4 π2 r / T2 .................. Proved
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Tangential Acceleration
The acceleration possessed by a body moving
along a circular path due to its changing speed during its motion is called
tangential acceleration. Its direction is along the tangent of the circular
path. It is denoted by a(t). If the speed is uniform (unchanging) the body do
not passes tangential acceleration.
Total Or Resultant Acceleration
The resultant of centripetal acceleration a(c)
and tangential acceleration a(t) is called total or resultant acceleration
denoted by a.
Centripetal Force
If a body is moving along a circular path with a
constant speed, a force must be acting upon it. Direction of the force is along
the radius towards the centre. This force is called the centripetal force by
F(c).
F(c) = m a(c)
F(c) = m v(2) / r ..................... {since
a(c) = v2 / r}
F(c) = mr2 ω2 r ....................... {since v
= r ω}
F(c) = mrω2
Motion
Definition
If an object continuously changes its position with respect to its
surrounding, then it is said to be in state of motion.
Rectilinear Motion
The motion along a straight line is called
rectilinear motion.
Velocity
Velocity may be defined as the change of
displacement of a body with respect time.
Velocity = change of displacement / time
Velocity is a vector quantity and its unit in
S.I system is meter per second (m/sec).
Average Velocity
Average velocity of a body is defined as the
ratio of the displacement in a certain direction to the time taken for this
displacement.
Suppose a body is moving along the path AC as
shown in figure. At time t1, suppose the body is at P and its position w.r.t
origin O is given by vector r2.
Diagram Coming Soon Thus, displacement of the
body = r2 - r1 = Δr
Time taken for this displacement - t2 - t1 = Δt
Therefore, average velocity of the body is given
by
Vav = Δr / Δt
Instantaneous Velocity
It is defined as the velocity of a body at a
certain instant.
V(ins) = 1im Δr / Δt
Where Δt → 0 is read as "Δt tends to zero", which
means that the time is very small.
Velocity From Distance - Time Graph
We can determine the velocity of a body by
distance - time graph such that the time is taken on x-axis and distance on
y-axis.
Acceleration
Acceleration of a body may be defined as the
time rate of change velocity. If the velocity of a body is changing then it is
said to posses acceleration.
Acceleration = change of velocity / time
If the velocity of a body is increasing, then its
acceleration will be positive and if the velocity of a body is decreasing, then
its acceleration will be negative. Negative acceleration is also called retardation.
Acceleration is a vector quantity and its unit
in S.I system is meter per second per second. (m/sec2 OR m.sec-2)
Average Acceleration
Average acceleration is defined as the ratio of
the change in velocity of a body and the time interval during which the
velocity has changed.
Suppose that at any time t1 a body is at A
having velocity V1. At a later time t2, it is at point B having velocity V2.
Thus,
Change in Velocity = V2 - V1 = Δ V
Time during which velocity has changed = t2 - t1
= Δ t
Instantaneous Acceleration
It is defined as the acceleration of a body at a
certain instant
a(ins) = lim Δ V / Δ t
where Δt → 0 is read as "Δt tends to zero", which
means that the time is very small.
Acceleration from Velocity - Time Graph
We can determine the acceleration of a body by
velocity - time graph such that the time is taken on x-axis and velocity on
y-axis.
Equations of Uniformly Accelerated Rectilinear Motion
There are three basic equations of motion. The
equations give relations between
Vi = the initial velocity of the body moving
along a straight line.
Vf = the final velocity of the body after a
certain time.
t = the time taken for the change of velocity
a = uniform acceleration in the direction of
initial velocity.
S = distance covered by the body.
Equations are
1. Vf = Vi + a t
2. S = V i t + 1/2 a t2
3. 2 a S = V f2 - V i 2
Motion Under Gravity
The force of attraction exerted by the earth on
a body is called gravity or pull of earth. The acceleration due to gravity is
produced in a freely falling body by the force of gravity. Equations for motion
under gravity are
1. Vf = Vi + g t
2. S = V i t + 1/2 g t2
3. 2 g S = Vr2 - Vi2
where g = 9.8 m / s2 in S.I system and is called
acceleration due to gravity.
Law of Motion
Isaac Newton studied motion of bodies and
formulated three famous laws of motion in his famous book "Mathematical
Principles of Natural Philosophy" in 1687. These laws are called Newton's
Laws of Motion.
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Newton's First Law of Motion
Statement
A body in state of rest will remain at rest and a body in state of motion
continues to move with uniform velocity unless acted upon by an unbalanced
force.
Explanation
This law consists of two parts. According to
first part a body at rest will remain at rest will remain at rest unless some
external unbalanced force acts on it. It is obvious from our daily life
experience. We observe that a book lying on a table will remain there unless
somebody moves it by applying certain force. According to the second part of
this law a body in state of uniform motion continuous to do so unless it is
acted upon by some unbalanced force.
This part of the law seems to be false from our
daily life experience. We observe that when a ball is rolled in a floor, after
covering certain distance, it stops. Newton gave reason for this stoppage that
force of gravity friction of the floor and air resistance are responsible of
this stoppage which are, of course, external forces. If these forces are not
present, the bodies, one set into motion, will continue to move for ever.
Qualitative Definition of Net Force
The first law of motion gives the qualitative
definition of the net force. (Force is an agent which changes or tends to
change the state of rest or of uniform motion of a body).
First Law as Law of Inertia
Newton's first law of motion is also called the
Law of inertia. Inertia is the property of matter by virtue of which is
preserves its state of rest or of uniform motion. Inertia of a body directly
related to its mass.
Newton's Second Law of Motion
Statement
If a certain unbalanced force acts upon a body, it produces acceleration in
its own direction. The magnitude of acceleration is directly proportional to
the magnitude of the force and inversely proportional to the mass of the body.
Mathematical Form
According to this law
f ∞ a
F = m a → Equation of second law
Where 'F' is the unbalanced force acting on the
body of mass 'm' and produces an acceleration 'a' in it.
From equation
1 N = 1 kg x 1 m/sec2
Hence one newton is that unbalanced force which produces
an acceleration of 1 m/sec2 in a body of mass 1 kg.
Vector Form
Equation of Newton's second law can be written
in vector form as
F = m a
Where F is the vector sum of all the forces
acting on the body.
Newton's Third Law of Motion
Statement
To every action there is always an equal and opposite reaction.
Explanation
For example, if a body A exerts force on body B
(F(A) on B) in the opposite direction. This force is called reaction. Then according to third law of motion.
Examples
1. When a gun is fired, the bullet flies out in
forward direction. As a reaction of this action, the gun reacts in backward
direction.
2. A boatman, when he wants to put his boat in
water pushes the bank with his oar, The reaction of the bank pushes the boat in
forward direction.
3. While walking on the ground, as an action, we
push the ground in the backward direction. As a reaction ground pushes us in
the forward direction.
4. In flying a kite, the string is given a
downward jerk and is then released. Thereupon the reaction of the air pushes
the kite upward and makes it rise higher.
Tension in a String
Consider a body of weight W supported by a
person with the help of a string. A force is experienced by the hand as well as
by the body. This force is known as Tension. At B the hand experiences a downward force. So the
direction of force at point B is downward. But at point A direction of the
force is upward.
These forces at point A and B are tensions. Its
magnitude in both cases is same but the direction is opposite. At point A,
Tension = T = W = mg
Momentum of a Body
The momentum of a body is the quantity of motion
in it. It depends on two things
1. The mass of the object moving (m),
2. The velocity with which it is moving (V).
Momentum is the product of mass and velocity. It
is denoted by P.
P = m V
Momentum is a vector quantity an its direction
is the same as that of the velocity.
Unit of Momentum
Momentum = mass x velocity
= kg x m/s
= kg x m/s x s/s
= kg x m/s2 x s
since kg. m/s2 is newton (N)
momentum = N-s
Hence the S.I unit of momentum is N-s.
Unbalanced or Net Force is equal to the Rate of Change of Momentum
i.e., F = (mVf = mVi) / t
Proof
Consider a body of mass 'm' moving with a
velocity Vl. A net force F acts on it for a time 't'. Its velocity then becomes
Vf.
Therefore
Initial momentum of the body = m Vi
Final momentum of the body = m Vf
Time interval = t
Unbalanced force = F
Therefore
Rate of change of momentum = (m Vf - m Vi) / t
....................... (1)
But
(Vf - Vi) / t = a
Therefore,
Rate of change of momentum = m a = F
..................... (2)
Substituting the value of rate of change of
momentum from equation (2) in equation (1), we get
F = (m Vf - m Vi) / t .............................
Proved
Law of Conservation of Momentum
Isolated System
When a number of bodies are such that they exert
force upon one another and no external agency exerts a force on them, then they
are said to constitute and isolated system.
Statement of the Law
The total momentum of an isolated system of bodies remains constant.
OR
If there is no external force applied to a system, then the total momentum
of that system remains constant.
Elastic Collision
An elastic collision is that in which the momentum
of the system as well as the kinetic energy of the system before and after
collision, remains constant. Thus for an elastic collision.
If P momentum and K.E is kinetic energy.
P(before collision) = P(after collision)
K.E(before collision) = K.E(after collision)
Inelastic Collision
An inelastic collision is that in which the
momentum of the system before and after the collision remains constant but the
kinetic energy before and after the collision changes.
Thus for an inelastic collision
P(before collision) = P(after collision)
Elastic Collision in one Dimension
Consider two smooth non rotating spheres moving
along the line joining their centres with velocities U1 and U2. U1 is greater
than U2, therefore the spheres of mass m1 makes elastic collision with the
sphere of mass m2. After collision, suppose their velocities become V1 and V2
but their direction of motion is along same line as before.
Friction
When two bodies are in contact, one upon the
other and a force is applied to the upper body to make it move over the surface
of the lower body, an opposing force is set up in the plane of the contract
which resists the motion. This force is the force of friction or simply
friction.
The force of friction always acts parallel to
the surface of contact and opposite to the direction of motion.
Definition
When one body is at rest in contact with another, the friction is called
Static Friction.
When one body is just on the point of sliding over the other, the friction
is called Limiting Friction.
When one body is actually sliding over the
other, the friction is called Dynamic Friction.
Coefficient of Friction (μ)
The ratio of limiting friction 'F' to the normal
reaction 'R' acting between two surfaces in contact is called the coefficient
of friction (μ).
μ = F / R
Or
F = μ R
Fluid Friction
Stoke found that bodies moving through fluids
(liquids and gases) experiences a retarding force fluid friction or viscous
drag. If the moving bodies are spheres then fluid friction F is given by
F = 6 π η r v
Where η is the coefficient of viscosity,
Where r is the radius of the sphere,
Where v is velocity pf the sphere.
Terminal Velocity
When the fluid friction is equal to the downward
force acting on the sphere, the sphere attains a uniform velocity. This
velocity is called Terminal velocity.
The Inclined Plane
A plane which makes certain angle θ with the
horizontal is called an inclined plane.
Diagram Coming Soon Consider a block of mass 'm'
placed on an inclined plane making certain angle θ with the horizontal. The
forces acting on the block are
1. W, weight of the block acting vertically
downward.
2. R, reaction of the plane acting perpendicular
to the plane
3. f, force of friction which opposes the motion
of the block which is moving downward.
Diagram Coming Soon Now we take x-axis along the
plane and y-axis perpendicular to the plane. We resolve W into its rectangular
components.
Therefore,
Component of W along x-axis = W sin θ
And
Component of W along y-axis = W cos θ
1. If the Block is at Rest
According to the first condition of equilibrium
Σ Fx = 0
Therefore,
f - W sin θ = 0
Or
f = W sin θ
Also,
Σ Fy = 0
Therefore,
R - W cos θ = 0
Or
R = W cos θ
2. If the Block Slides Down the Inclined Plane with an Acceleration
Therefore,
W sin θ > f
Net force = F = W sin θ - f
Since F = m a and W = m g
Therefore,
m a = m g sin θ - f
3. When force of Friction is Negligible
Then f ≈ 0
Therefore,
equation (3) => m a = m g sin ≈ - 0
=> m a = m g sin ≈
or a = g sin ≈ ............. (4)
Particular Cases
Case A : If the Smooth Plane is Horizontal Then 0 = 0º
Therefore,
Equation (4) => a = g sin 0º
=> a = g x 0
=> a = 0
Case B : If the Smooth Plane is Vertical Then θ = 90º
Therefore,
Equation (4) => a = g sin 90º
=> a = g x 1
=> a = g
This is the case of a freely falling body.
Scalars and
Vectors
Scalars
Physical quantities which can be completely
specified by
1. A number which represents the magnitude of
the quantity.
2. An appropriate unit
are called Scalars.
Scalars quantities can be added, subtracted
multiplied and divided by usual algebraic laws.
Examples
Mass, distance, volume, density, time, speed,
temperature, energy, work, potential, entropy, charge etc.
Vectors
Physical quantities which can be completely
specified by
1. A number which represents the magnitude of
the quantity.
2. An specific direction
are called Vectors.
Special laws are employed for their mutual
operation.
Examples
Displacement, force, velocity, acceleration,
momentum.
Representation of a Vector
A straight line parallel to the direction of the
given vector used to represent it. Length of the line on a certain scale
specifies the magnitude of the vector. An arrow head is put at one end of the
line to indicate the direction of the given vector.
The tail end O is regarded as initial point of vector R and the head P is regarded as the terminal point of the vector R.
Diagram Coming Soon
Unit Vector
A vector whose magnitude is unity (1) and
directed along the direction of a given vector, is called the unit vector of
the given vector.
A unit vector is usually denoted by a letter
with a cap over it. For example if r is the given vector, then r will be the
unit vector in the direction of r such that
r = r .r
Or
r = r / r
unit vector = vector / magnitude of the vector
Equal Vectors
Two vectors having same directions, magnitude
and unit are called equal vectors.
Zero or Null Vector
A vector having zero magnitude and whose initial
and terminal points are same is called a null vector. It is usually denoted by
O. The difference of two equal vectors (same vector) is represented by a null
vector.
R - R - O
Free Vector
A vector which can be displaced parallel to
itself and applied at any point, is known as free vector. It can be specified
by giving its magnitude and any two of the angles between the vector and the
coordinate axes. In 3-D, it is determined by its three projections on x, y,
z-axes.
Position Vector
A vector drawn from the origin to a distinct
point in space is called position vector, since it determines the position of a
point P relative to a fixed point O (origin). It is usually denoted by r. If
xi, yi, zk be the x, y, z components of the position vector r, then
r = xi + yj + zk
Diagram Coming Soon
Negative of a Vector
The vector A. is called the negative of the
vector A, if it has same magnitude but opposite direction as that of A. The
angle between a vector and its negative vector is always of 180º.
Multiplication of a Vector by a Number
When a vector is multiplied by a positive number
the magnitude of the vector is multiplied by that number. However, direction of
the vector remain same. When a vector is multiplied by a negative number, the
magnitude of the vector is multiplied by that number. However, direction of a
vector becomes opposite. If a vector is multiplied by zero, the result will be
a null vector.
The multiplication of a vector A by two number
(m, n) is governed by the following rules.
1. m A = A m
2. m (n A) = (mn) A
3. (m + n) A = mA + nA
4. m(A + B) = mA + mB
Division of a Vector by a Number (Non-Zero)
If a vector A is divided by a number n, then it
means it is multiplied by the reciprocal of that number i.e. 1/n. The new
vector which is obtained by this division has a magnitude 1/n times of A. The
direction will be same if n is positive and the direction will be opposite if n
is negative.
Resolution of a Vector Into Rectangular Components
Definition
Splitting up a single vector into its
rectangular components is called the Resolution of a vector.
Rectangular Components
Components of a vector making an angle of 90º
with each other are called rectangular components.
Procedure
Let us consider a vector F represented by OA,
making an angle O with the horizontal direction.
Draw perpendicular AB and AC from point on X and
Y axes respectively. Vectors OB and OC represented by Fx and Fy are known as
the rectangular components of F. From head to tail rule of vector addition.
OA = OB + BA
F = Fx + Fy
Diagram Coming Soon To find the magnitude of Fx
and Fy, consider the right angled triangle OBA.
Fx / F = Cos θ => Fx = F cos θ
Fy / F = sin θ => Fy = F sin θ
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Addition of Vectors by Rectangular Components
Consider two vectors A1 and A2 making angles θ1
and θ2 with x-axis respectively as shown in figure. A1 and A2 are added by
using head to tail rule to give the resultant vector A.
Diagram Coming Soon The addition of two vectors
A1 and A2 mentioned in the above figure, consists of following four steps.
Step 1
For the x-components of A, we add the
x-components of A1 and A2 which are A1x and A2x. If the x-components of A is
denoted by Ax then
Ax = A1x + A2x
Taking magnitudes only
Ax = A1x + A2x
Or
Ax = A1 cos θ1 + A2 cos θ2 ................. (1)
Step 2
For the y-components of A, we add the
y-components of A1 and A2 which are A1y and A2y. If the y-components of A is
denoted by Ay then
Ay = A1y + A2y
Taking magnitudes only
Ay = A1y + A2y
Or
Ay = A1 sin θ1 + A2 sin θ2 ................. (2)
Step 3
Substituting the value of Ax and Ay from
equations (1) and (2) respectively in equation (3) below, we get the magnitude
of the resultant A
A = |A| = √ (Ax)2 + (Ay)2 .................. (3)
Step 4
By applying the trigonometric ratio of tangent θ
on triangle OAB, we can find the direction of the resultant vector A i.e. angle
θ which A makes with the positive x-axis.
tan θ = Ay / Ax
θ = tan-1 [Ay / Ax]
Here four cases arise
(a) If Ax and Ay are both positive, then
θ = tan-1 |Ay / Ax|
(b) If Ax is negative and Ay is positive, then
θ = 180º - tan-1 |Ay / Ax|
(c) If Ax is positive and Ay is negative, then
θ = 360º - tan-1 |Ay / Ax|
(d) If Ax and Ay are both negative, then
θ = 180º + tan-1 |Ay / Ax|
Addition of Vectors by Law of Parallelogram
According to the law of parallelogram of
addition of vectors, if we are given two vectors. A1 and A2 starting at a common
point O, represented by OA and OB respectively in figure, then their resultant
is represented by OC, where OC is the diagonal of the parallelogram having OA
and OB as its adjacent sides.
Diagram Coming Soon If R is the resultant of A1
and A2, then
R = A1 + A2
Or
OC = OA + OB
But OB = AC
Therefore,
OC = OA + AC
β is the angle opposites to the resultant.
Magnitude of the resultant can be determined by
using the law of cosines.
R = |R| = √A1(2) + A2(2) - 2 A1 A2 cos β
Direction of R can be determined by using the
Law of sines.
A1 / sin γ = A2 / sin α = R / sin β
This completely determines the resultant vector
R.
Properties of Vector Addition
1. Commutative Law of Vector Addition (A+B = B+A)
Consider two vectors A and B as shown in figure.
From figure
OA + AC = OC
Or
A + B = R .................... (1)
And
OB + BC = OC
Or
B + A = R ..................... (2)
Since A + B and B + A, both equal to R,
therefore
A + B = B + A
Therefore, vector addition is commutative.
Diagram Coming Soon
2. Associative Law of Vector Addition (A + B) + C = A + (B + C)
Consider three vectors A, B and C as shown in
figure. From figure using head - to - tail rule.
OQ + QS = OS
Or
(A + B) + C = R
And
OP + PS = OS
Or
A + (B + C) = R
Hence
(A + B) + C = A + (B + C)
Therefore, vector addition is associative.
Diagram Coming Soon
Product of Two Vectors
1. Scalar Product (Dot Product)
2. Vector Product (Cross Product)
1. Scalar Product OR Dot Product
If the product of two vectors is a scalar
quantity, then the product itself is known as Scalar Product or Dot Product.
The dot product of two vectors A and B having
angle θ between them may be defined as the product of magnitudes of A and B and
the cosine of the angle θ.
A . B = |A| |B| cos θ
A . B = A B cos θ
Diagram Coming Soon Because a dot (.) is used
between the vectors to write their scalar product, therefore, it is also called
dot product.
The scalar product of vector A and vector B is
equal to the magnitude, A, of vector A times the projection of vector B onto
the direction of A.
If B(A) is the projection of vector B onto the
direction of A, then according to the definition of dot product.
Diagram Coming Soon A . B = A B(A)
A . B = A B cos θ {since B(A) = B cos θ}
Examples of dot product are
W = F . d
P = F . V
Commutative Law for Dot Product (A.B = B.A)
If the order of two vectors are changed then it
will not affect the dot product. This law is known as commutative law for dot
product.
A . B = B . A
if A and B are two vectors having an angle θ
between then, then their dot product A.B is the product of magnitude of A, A, and the projection of vector B onto the direction of vector i.e.,
B(A).
And B.A is the product of magnitude of B, B, and the projection of vector A onto the direction vector B i.e. A(B).
Diagram Coming Soon To obtain the projection of
a vector on the other, a perpendicular is dropped from the first vector on the
second such that a right angled triangle is obtained
In Δ PQR,
cos θ = A(B) / A => A(B) = A cos θ
In Δ ABC,
cos θ = B(A) / B => B(A) = B cos θ
Therefore,
A . B = A B(A) = A B cos θ
B . A = B A (B) = B A cos θ
A B cos θ = B A cos θ
A . B = B . A
Thus scalar product is commutative.
Distributive Law for Dot Product
A . (B + C) = A . B + A . C
Consider three vectors A, B and C.
B(A) = Projection of B on A
C(A) = Projection of C on A
(B + C)A = Projection of (B + C) on A
Therefore
A . (B + C) = A [(B + C}A] {since A . B = A
B(A)}
= A [B(A) + C(A)] {since (B + C)A = B(A) + C(A)}
= A B(A) + A C(A)
= A . B + A . C
Therefore,
B(A) = B cos θ => A B(A) = A B cos θ1 = A . B
And C(A) = C cos θ => A C(A) = A C cos θ2 = A
. C
Thus dot product obeys distributive law.
Diagram Coming Soon
2. Vector Product OR Cross Product
When the product of two vectors is another
vector perpendicular to the plane formed by the multiplying vectors, the
product is then called vector or cross product.
The cross product of two vector A and B having
angle θ between them may be defined as "the product of magnitude of A and
B and the sine of the angle θ, such that the product vector has a direction
perpendicular to the plane containing A and B and points in the direction in
which right handed screw advances when it is rotated from A to B through
smaller angle between the positive direction of A and B".
A x B = |A| |B| sin θ u
Where u is the unit vector perpendicular to the
plane containing A and B and points in the direction in which right handed
screw advances when it is rotated from A to B through smaller angle between the
positive direction of A and B.
Examples of vector products are
(a) The moment M of a force about a point O is
defined as
M = R x F
Where R is a vector joining the point O to the
initial point of F.
(b) Force experienced F by an electric charge q
which is moving with velocity V in a magnetic field B
F = q (V x B)
Physical Interpretation of Vector OR Cross Product
Area of Parallelogram = |A x B|
Area of Triangle = 1/2 |A x B|