Work, Energy And Power ( Physics ) MCQs – Physics MCQs
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Physics MCQs – Work, Energy And Power ( Physics ) MCQs
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Work-Energy Theorem for a Variable Force
1. For a body moving from position Xa to position Xb under a variable force, what is the work done?
a) W = F * (X2 – X1)
b) -W = F * (X2 – X1)
c) W = -F * (X2 – X1)
d) W = F * (X1 – X2)
Answer: a
Explanation: Kinetic energy (K) = 1/2 x m x v2
dK/dt = m*v*dv/dt
Force (F) = m*a
= m*dv/dt
dK/dt = F*v
= m*v*dv/dt
∫K2K1 dK/dt = ∫XbXa F * dX/dt
∫K2K1 dK = ∫XbXa F * dX
K2 – K1 = F (X2 – X1)
W = F * (X2 – X1).
2. The work-energy theorem is valid only for positive work done.
a) True
b) False
Answer: b
Explanation: Work energy theorem holds for both positive and negative work done. If the work doneis positive then final kinetic energy increases by the amount of the work and if work done is negative then final kinetic energy decreases by the amount of work done.
3. A car having a mass of 1 metric ton is moving with a speed of 30 m/s. It suddenly applies the brakes and skids to rest in a certain distance d. The frictional force between the tyres and road is 6000 N. What is the value of d?
a) 25 m
b) 50 m
c) 75 m
d) 100 m
Answer: c
Explanation: Let; m = 1000 kg
From Work-Energy theorem
Kf – Ki = F * d
0 – (1/2 * m * v2) = -6000 * d
1/2 x 1000 x 302 = 6000 x d
d = 75 m.
4. For the graph shown in the figure, what is the difference between x-coordinate of A and B called?
a) Internal pressure
b) External pressure
c) Work
d) Power
Answer: c
Explanation: The difference between x-coordinate of A and B can be written as “Kb – Ka”. This is the difference between kinetic energies at two positions. This gives us the “work done”.
5. For the graph shown in the figure, what is the area under the curve AB called?
a) Force applied
b) Displacement
c) Work
d) Power
Answer: c
Explanation: The area under the curve is a summation of the product of x-coordinate and y-coordinate at every point under the curve.
That is; “F * x”.
F * x = Work done.
6. A “variable force” is the force that only increases with time.
a) True
b) False
Answer: b
Explanation: A “variable force” is the force that changes with time. This force can increase or decrease at random time intervals. However, it is not constant with respect to time.
i.e., dF/dt is not equal to “zero”.
Potential Energy
1. The potential energy of an object maximises as its velocity increases.
a) True
b) False
Answer: b
Explanation: The potential velocity of an object is maximum when it is at rest. The kinetic energy increases as velocity increases while the potential energy decreases.
2. An object of mass 5kg is taken from a height of 10 m to a height of 30 m. What is the increment in its potential energy? (Assume g = 10m/s2)
a) 500 J
b) 1000 J
c) 1500 J
d) 2000 J
Answer: b
Explanation: PE = m x g x h
m = 5 kg
g = 10 m/s2
Let;
h1 = 10m
h2 = 30m
PE1 = 5 x 10 x 10
= 500 J
PE2 = 5 x 10 x 30
= 1500 J
Increment in PE = PE2 – PE1
= 1500 – 500
= 1000 J.
3. A 10kg object is raised to a height of 20m. What is the magnitude of its potential energy? (Assume g = 10 m/s2)
a) 500 J
b) 1000 J
c) 1500 J
d) 2000 J
Answer: d
Explanation: PE = m x g x h
m = 10 kg
g = 10 m/s2
h = 20 m
PE = 10 x 10 x 20
= 2000 J.
4. An object is thrown from the ground with a velocity of 5 m/s. If the object has a mass of 2kg, what will be its potential energy at the top-most point of its trajectory? (Assume g = 10 m/s2)
a) 5 J
b) 15 J
c) 25 J
d) 50 J
Answer: c
Explanation: Maximum height reached by the object;
h = v2/(2 x g)
= 25/20
= 1.25 m
PE = m x g x h
= 2 x 10 x 1.25
= 25 J.
5. Which of the following is true?
a) Potential energy decreases as altitude increases
b) Potential energy increases as altitude increases
c) Potential energy first increases and then decreases as altitude increases
d) Potential energy first decreases and then increases as altitude increases
Answer: b
Explanation: As altitude increases potential energy increases. From the equation of potential energy (PE);
PE = m x g x h
We can see that the potential energy is directly proportional to height. Hence the statement is justified.
6. Since the potential energy of an object depends on the acceleration due to gravity and since the acceleration due to gravity decreases as altitude increases, we can conclude that the potential energy of an object decreases as altitude increases.
a) True
b) False
Answer: b
Explanation: The potential energy of an object does not depend on acceleration due to gravity at the instantaneous position of the object but depends on the acceleration due to gravity at the reference point (such as the surface of the earth). Hence, potential energy increases as altitude increases.
7. Consider 2 balls A and B of the same mass. The potential energy of ball A is thrice that of ball B. How high is ball A compared to ball B?
a) Same height as ball B
b) Twice as high height as ball B
c) Thrice as high height as ball B
d) Four times as high height as ball B
Answer: c
Explanation: PEa = (m x g x h) a
PEb = (m x g x h) b
ma = mb = m
ga = gb = g
PEa = 3 x PEb
ma x ga x ha = mb x gb x hb x 3
m x g x ha = m x g x hb x 3
ha = 3 x hb.
8. The potential energy of an object of constant mass and fixed reference is determined by its _____
a) mass
b) gravitational acceleration
c) position
d) velocity
Answer: c
Explanation: PE = m x g x h
The mass of the given object is constant.
The fixed reference indicates constant acceleration due to gravity.
Hence, variation in height (h) is nothing but a variation in position.
Hence, the potential energy of an object of constant mass and fixed reference is determined by its position.
9. The potential energy of an object cannot be increased by internal forces.
a) True
b) False
Answer: a
Explanation: The potential energy can only be increased by external forces. Energy increase by internal forces manifests as an increase in kinetic energy. However, potential energy can be reduced by internal forces by increasing kinetic energy.
10. Find the ratio of potential energy if an object is raised to thrice of its height and its mass is tripled.
a) 1:1
b) 1:4
c) 1:9
d) 1:3
Answer: c
Explanation: PE1 = m x g x h
PE2 = (3m) x g x (3h)
= 9 x m x g x h
PE1 : PE2 = mgh : 9mgh
= 1:9.
The Potential Energy of a Spring
1. The potential energy possessed by a spring is also known as _____
a) Elastic potential energy
b) Extensive potential energy
c) Compressive potential energy
d) Deflection potential energy
Answer: a
Explanation: The potential energy stored in a spring is the consequence of its elastic property and hence it is also termed “elastic potential energy”. It can be stored when the spring is extended or compressed.
2. An object is travelling at a velocity of 10 m/s. It has a mass of 2 kg. It impacts a spring of spring constant “k” as shown in the figure. What is the compression of the spring?
a) 10 x (2/k) 0.5
b) 10 x (200/k) 0.5
c) (200/k)
d) (200/k) 2
Answer: a
Explanation: From the law of conservation of energy;
1/2 x m x v2 = 1/2 x k x d2
2 x 102 = k x d2
200/k = d2
d = (200/k) 0.5
= 10 x (2/k) 0.5.
3. How much should a spring of indefinite length be compressed to have a potential energy equivalent to a ball of mass 6 kg raised to a height of 120 m above the ground? Let the spring have a stiffness of k = 100 N/m and assume g = 10 m/s2.
a) 6m
b) 12m
c) 20m
d) 144m
Answer: b
Explanation: PE of ball = m x g x h
= 6 x 10 x 120
= 7200 J
PE of spring = 1/2 x k x d2
Given;
PE of ball = PE of spring
7200 = 1/2x k x d2
= 1/2 x 100 x d2
d2 = 144
d = 12 m.
4. The following graph represents theparabolic plots of the potential energy V andkinetic energy K ofa block attached to aspring obeying Hooke’s law.
a) True
b) False
Answer: a
Explanation: The following graph represents the parabolic plots of the potential energy V and kinetic energy K of a block attached to a spring obeying Hooke’s law.The two plotsare complementary, onedecreasing as theother increases. The total mechanicalenergy E = K + V remains constant.
5. Find the energy equation for the system shown in the figure. “m” is the mass of the block moving at a velocity “v”, “k” is the spring constant, “q” is the coefficient of friction of the floor and let “d” be the compression in the spring.
a) 1/2*m*v2+ = 1/2*k*d2 + q*m*g*d
b) 1/2*m*v2+ + 1/2*k*d2 + q*m*g*d = 0
c) -1/2*m*v2+ = 1/2*k*d2 + q*m*g*d
d) 1/2*m*v2+ – 1/2*k*d2 + q*m*g*d = 0
Answer: a
Explanation: Work done by velocity (W1) = 1/2 x m x v2
Work done by spring (W2) = 1/2 x k x d2
Work done by friction (W3) = q x m x g x d
Since the work done by spring and friction are opposite to that of the work done by the velocity, we have;
W1 – W2 – W3 = 0
W1 = W2 + W3
Therefore;
1/2*m*v2+ = 1/2*k*d2 + q*m*g*d.
6. Assume a spring extend by “d” due to some load. Let “F” be the spring force and “k’ the spring constant. Then, the potential energy stored is _____
a) 2d/F2
b) F2/2k
c) 2k/T2
d) F2/2d
Answer: b
Explanation: F = k*d
PE = 1/2*k*d2
= 1/2*F*d
= 1/2*F*F/k
= F2/2k.
7. A spring of length 1m has two cars connected to both of its ends. The two cars move towards eachother such that the spring is compressed to 0.5m. If the spring constant is 500 N/m, what is the elastic potential energy stored?
a) 125 J
b) -125 J
c) 62.5 J
d) -62.5 J
Answer: c
Explanation: PE = 1/2*k*d2
= 1/2 x 500 x (-0.5) 2
= 62.5 J.
The elastic potential energy is positive even for negative displacement.
8. What is the increase in potential energy storage when the compression distance is doubled in a spring obeying Hooke’s law?
a) No increase
b) 100% increase
c) Cannot be determined
d) 4 times
Answer: d
Explanation: PE = 1/2 x k x d2
The elastic potential energy stored is directly proportional to the square of the compression. Hence, a two-fold increase in compression will result in potential energy storage to increase by four times.
9. The elastic potential energy varies linearly with displacement.
a) True
b) False
Answer: b
Explanation: PE = 1/2 x k x d2; where d = Displacement
Equation of parabola: y = 4x2
By comparing the two equations, we can conclude that the elastic potential energy varies parabolically with displacement and not linearly.
Various Forms of Energy : The Law of Conservation of Energy
1. Which of the following statements are true?
a) Energy can be created and destroyed
b) Energy cannot be created but only destroyed
c) Energy cannot be destroyed but only created
d) Energy can neither be created nor destroyed
Answer: d
Explanation: According to the law of conservation of energy, energy can neither be created nor destroyed but can only be changed from one form to another. This is also one of the fundamental concepts of thermodynamics.
2. The dimensions of energy are _____
a) [ML2T-2]
b) [MLT-2]
c) [ML2T-1]
d) [MLT-1]
Answer: a
Explanation: Energy = Force x Displacement
Units of energy = kg x (m/s2) x m
= kg x m2 x s-2
Therefore;
Dimensions of energy = [ML2T-2].
3. Which of the following is not true?
a) Energy of an open system can be varied
b) Energy cannot be created
c) Energy cannot be destroyed
d) Energy cannot be transformed
Answer: d
Explanation: According to the law of conservation of energy, energy can neither be created nor destroyed but can only be changed from one form to another. Energy in an open system can be varying because of constant matter and radiation exchanges with the surrounding universe.
4. The law of conservation of energy is not applicable to mechanical systems as they require energy input to keep working.
a) True
b) False
Answer: b
Explanation: The law of conservation of energy is applicable to mechanical systems but it may not necessarily be applicable to a “machine” that requires constant power input to operate because it would then essentially be an open system.
5. The energy possessed by an object because of its motion is termed _____
a) potential energy
b) kinetic energy
c) nuclear energy
d) solar energy
Answer: b
Explanation: An object in motion possesses a velocity “v” at any instant of time.
This energy is termed “kinetic energy” and is given by;
Kinetic Energy = 1/2 x m x v2
where;
m = Mass of the object in motion
6. The maximum potential energy in a roller coaster is at _____
a) the top of the steep climb
b) somewhere during the climb
c) somewhere during the descent
d) the lowest point after the climb
Answer: a
Explanation: The potential energy is given by;
PE = m x g x h
where;
PE = Potential energy
m = Mass of the object
g = Acceleration due to gravity
h = Height of the object
Hence, the higher the object, the greater the potential energy.
7. Fire is a form of _____
a) solar energy
b) thermal energy
c) gravitational energy
d) kinetic energy
Answer: b
Explanation: Thermal energy is the consequence of the temperature of an object that is transferred in the form of heat from a source of greater temperature to a region or object of lower temperature by modes such as conduction, convection or radiation.
8. When a roller coaster is at the top of a steep hill, its potential energy is maximum. When it hurdles down the slope and comes to the bottom, the potential energy reduces greatly. Hence we can say that the law of conservation of energy is not valid here since energy is destroyed.
a) True
b) False
Answer: b
Explanation: When a roller coaster is at the top of a steep hill, its potential energy is maximum. When it hurdles down the slope and comes to the bottom, the potential energy reduces greatly. However, kinetic energy increases. Thus, there is a change from potential energy to kinetic energy and hence no destruction of energy.
9. When a person intakes 3000 kcal of food, he is able to do work requiring energy of 2500 kcal. This means that the human body destroys a fraction of energy obtained from food.
a) True
b) False
Answer: b
Explanation: If a person consumes 3000 kcal of food and does a work requiring 2500 kcal of energy, the rest of the 500 kcal is mostly used for the metabolic activities of the body and some are lost in the form of heat energy to the atmosphere.
10. For a freely falling body, which of the following quantities will not change?
a) Total kinetic energy
b) Total potential energy
c) Total mechanical energy
d) Insufficient data
Answer: c
Explanation: For a body falling freely under the influence of gravity, the velocity increases as heigh decreases. Hence, there is a reduction in potential energy but an increase in kinetic energy. However, the total mechanical energy, i.e., the sum of potential and kinetic energy always remains constant as per the law of conservation of energy.
Power
1. What are the units of power?
a) Newton
b) Joule
c) Watt
d) No units
Answer: c
Explanation: The SI unit of power is “Watt”. It is named after the famous inventor “James Watt” who is widely remembered for his improvements on the steam engine.
1 Watt = 1 Joule per second.
2. A machine gun fires 360 bullets per minute. Each bullet has a mass of 5 grams and travels at 600 m/s. What is the power of the gun? (Assume no loss of energy and 100% power transmission)
a) 300 W
b) 600 W
c) 900 W
d) 1800 W
Answer: c
Explanation: No. of bullets fired per second = 360/60 = 6
Power of the gun is completely transmitted to bullets.
Power (P) = Power of bullets in 1 second
= [(1/2 x m x v2) x 6] / 1
m = 5 x 10-3 kg
v = 600 m/s
P = (1/2 x 5 x 10-3 x 6002 x 6) / 1
= 900 W.
3. A motor is used to deliver water through a pipe. Let the motor have a power P initially. To double the rate of flow of water through the pipe, the power is increased to P’. What is the value of P’/P?
a) 2
b) 4
c) 6
d) 8
Answer: d
Explanation: Power = Force x Velocity
= (Mass/Time) x Velocity2
= Rate of flow x Velocity2
Rate of flow = Area x Velocity
Doubling the rate of flow will double the velocity.
P = Rate of flow x Velocity2
P’ = (2 x Rate of flow) x (2 x Velocity) 2
= 8P
P/P’ = 8.
4. A motor is used to pump water from a depth of 5 m to fill a volume of 10 cubic meters in 5 minutes. If 50% of the power is wasted, what is the power of the motor? (Assume g = 10m/s2)
a) 1000/3 W
b) 3000/3 W
c) 5000/3 W
d) 10000/3 W
Answer: d
Explanation: Density of water = 1000 kg/m3
Mass of water in 10m3 = 1000 x 10
= 10,000 kg
Volume of water filled in 1s = 10/(5×60)
= 1/30 m3
Mass filled in 1s = 1/30 x 1000
= 100/3 kg
Power = 100/3 x 10 x 5; [g = 10m/s2]
= 5000/3 W
Since 50% of power is wasted, power of motor = 5000/3 x 2
= 10000/3 W.
5. A machine has a power of 20kW. How long will it take for it to lift a body of mass 10kg from the ground to a height of 100m? (Assume g = 10m/s2)
a) 1 second
b) 2 seconds
c) 3 seconds
d) 4 seconds
Answer: b
Explanation: Power (P) = Energy (E) x Time (t)
20,000 = (m x g x h) x t
= (10 x 10 x 100) x t
t = 2 seconds.
6. A 2-ton vehicle travels at 20m/s on a road where the frictional force is 10% of the weight of the vehicle. What is the power required? (Assume g = 10m/s2)
a) 220 kW
b) 240 kW
c) 420 kW
d) 440 kW
Answer: d
Explanation:Weight of car = m x g
= 2,000 x 10
= 20,000 N
Frictional force = (10/100) x (20,000)
= 2,000 N
Total force (F) = 20,000 + 2,000
= 22,000 N
Velocity (v) = 20 m/s
P = F x v
= 22,000 x 20
= 4,40,000 W
= 440 kW.
7. 1000 kW of power is supplied to a motor. 90% of this is transmitted to a machine operating at 80% efficiency. The machine lifts an object of 10-tons with a velocity _____ m/s. (Assume g = 10m/s2)
a) 2
b) 4
c) 6
d) 8
Answer: c
Explanation: Power transmitted to machine = (90/100) x (1000) kW
= 900 kW
Power used to lift object = (80/100) x 900 KW
= 720 kW
7,20,000 = F x v
= (1/2 x m x v2) x v
= 10,000/2 x v3
v3 = 216
v = 6 m/s.
8. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 10 m/s. How much power is delivered by the engine? (Assume g = 10m/s2)
a) 10 kW
b) 20 kW
c) 30 kW
d) 40 kW
Answer: b
Explanation: Weight of car = m x g
= 2,00 x 10
= 2,000 N
Total force (F) = 2,000
Velocity (v) = 10 m/s
P = F x v
= 2000 x 10
= 20,000 W
= 20 kW.
9. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 36 km/h. How much power is delivered by the engine if the frictional force is 25% of the weight of the car? (Assume g = 10m/s2)
a) 25 kW
b) 50 kW
c) 100 kW
d)125 kW
Answer: a
Explanation: Weight of car = m x g
= 200 x 10
= 2,000 N
Frictional force = (25/100) x (2,000)
= 500 N
Total force (F) = 2000 + 500
= 2,500 N
Velocity (v) = 36 km/h
= 36 x (1000/3600)
= 10 m/s
P = F x v
= 2500 x 10
= 25 kW.
10. A body of mass 3 kg starts from rest with a uniform acceleration of unknown magnitude. If the body has a velocity of 30 m/s in 6 seconds, what is the power consumed in 3 seconds?
a) 125 W
b) 225 W
c) 325 W
d) 425 W
Answer: b
Explanation: v = u + at
30 = a x 6
a = 5 m/s2
Force (F) = m x a
= 3 x 5
= 15 N
v’ = u + at’
v’ = 5 x 3
= 15 m/s
Power consumed in 3 seconds = F x v’
= 15 X 15
= 225 W.
11. The force required to tow a vehicle at constant velocity is directly proportional to the magnitude of velocity raised to the first power. It requires 1000 W to tow with a velocity of 10 m/s. How much power is required to tow at a velocity of 8 m/s?
a) 320 W
b) 640 W
c) 160 W
d) 80 W
Answer: b
Explanation: Since force is directly proportional to velocity, the power required will be directly proportional to the square of the velocity.
Power = Force x Velocity
Hence, power is a multiple of 82 and an integer such that;
1000/102 = P/82
P = 640 W.
Collision
1. What is the coefficient of restitution (e)?
a) Ratio of final to initial relative velocities
b) Ratio of initial to final relative velocities
c) Ratio of final to initial velocity of a body
d) Product of initial to final relative velocities
Answer: a
Explanation: The coefficient of restitution (e) is the ratio of the final to initial relative velocities in a collision system. If the value of e = 1, then the collision is perfectly elastic. If it is equal to 0, then the collision is perfectly inelastic.
2. For what value of e is a collision perfectly elastic?
a) 1
b) 2
c) 0.5
d) 0
Answer: a
Explanation: A perfectly elastic collision occurs when the value of e is 1. When the value of e is less than one the collision starts to becomes inelastic.
3. For what value of e is a collision perfectly inelastic?
a) 1
b) 2
c) 0.5
d) 0
Answer: d
Explanation: A perfectly inelastic collision occurs when the value of e is 0. When the value of e is more than 0 the collision starts to becomes elastic.
4. For what value of e is a collision partially elastic?
a) 0 < e < 1
b) 1
c) e > 1
d) 0
Answer: a
Explanation: Partially elastic collision occurs when the value of e is less than one and more than zero. In this case the collision is neither perfectly elastic nor perfectly inelastic. Hence, we call it partially elastic collision.
5. A ball of mass 100 g is moving with a velocity of 3 m/s towards east. A cube of mass 200 gram is moving towards west with a velocity of 7 m/s. If the collision is perfectly elastic, what is the magnitude of the final momentum of the system in Kg-m/s?
a) 1700
b) 1500
c) 1300
d) 1100
Answer: d
Explanation: The initial momentum of the system = momentum of ball + momentum of cube = 1100 Kg-m/s. Since the momentum is conserved, the initial momentum = final momentum. Hence the final momentum is 1100 Kg-m/s.
6. A ball of mass 100 g is moving with a velocity of 3 m/s towards east. A cube of mass 200 gram is moving towards west with a velocity of 7 m/s. If the collision is perfectly elastic, what is the final velocity of the cube in m/s?
a) 7
b) 5
c) 3
d) 11
Answer: a
Explanation: The initial momentum of the system = momentum of ball + momentum of cube = 1100 Kg-m/s eastwards. Since the momentum is conserved, the initial momentum = final momentum. Hence the final momentum is 1100 Kg-m/s eastwards. The collision is perfectly elastic, therefore, e = 1. Therefore, final relative velocity (of cube with respect to ball) = initial relative velocity (of cube with respect to ball). Difference in the final velocities of cube and ball = 4 m/s. From this information and the conservation of momentum, one can find the final velocity of the cube.
7. A ball of mass 100 g is moving with a velocity of 3 m/s towards east. A cube of mass 200 gram is moving towards west with a velocity of 7 m/s. If the collision is perfectly elastic, what is the final velocity of the ball in m/s?
a) 7
b) 5
c) 31/3
d) 11/3
Answer: c
Explanation: The initial momentum of the system = momentum of ball + momentum of cube = 1100 Kg-m/s eastwards. Since the momentum is conserved, the initial momentum = final momentum. Hence the final momentum is 1100 Kg-m/s eastwards. The collision is perfectly elastic, therefore, e = 1. Therefore, final relative velocity (of cube with respect to ball) = initial relative velocity (of cube with respect to ball). Sum of the final velocities of cube and ball = 10 m/s. From this information and the conservation of momentum, one can find the final velocity of the ball.
8. A car of mass 1ton is moving with a velocity of 25 m/s towards left. A truck of mass 5 ton is moving towards left with a velocity of 3 m/s. If the collision is perfectly elastic, what is the final velocity of the truck in m/s?
a) 11
b) 25
c) 3
d) 10
Answer: c
Explanation: The initial momentum of the system = momentum of car + momentum of truck = 40 Kg-m/s leftwards. Since the momentum is conserved, the initial momentum = final momentum. Hence the final momentum is 40 Kg-m/s leftwards. The collision is perfectly elastic, therefore, e = 1. Therefore, final relative velocity (of truck with respect to car) = initial relative velocity (of truck with respect to car). Sum of the final velocities of the car and the truck = 4 m/s. From this information and the conservation of momentum, one can find the final velocity of the truck.
9. When two same masses travelling in opposite directions with different velocities collide perfectly elastically, their velocities ______
a) Exchange
b) Remain unchanged
c) Increase
d) Decrease
Answer: a
Explanation: In this case also we need to conserve the momentum and write the expression for e. After doing so we will find that the sum of the initial velocities is equal to the sum of the final velocities. Same goes for the difference (with slight alterations). This will result in the final velocity of one body being equal to the initial velocity of the other. Hence, the velocities will get exchanged.
10. A ball traveling with some velocity v hits another ball elastically which is stationary. What will happen to the system?
a) The stationary ball will start moving with a velocity v and the moving ball will come at rest
b) The stationary ball remains at rest and the moving ball starts moving in opposite direction
c) Both the balls become stationary
d) Both the balls start moving
Answer: a
Explanation: In this case also we need to conserve the momentum and write the expression for e. After some calculation we will find that the stationary ball will start moving with a velocity v and the moving ball will come at rest.
11. When e < 1, the kinetic energy in the collision is ____
a) Lost
b) Gained
c) Gained exponentially
d) Unchanged
Answer: a
Explanation: Coefficient of restitution or e can be understood as the measure of how much kinetic energy is remaining after the collision. As e becomes lesser and lesser, more kinetic energy is lost in the collision. This can be easily understood with by deriving the coefficient of restitution in a completely elastic collision.
12. Two bodies collide perfectly inelastically, they will ______
a) Move together
b) Move in opposite directions
c) Move separately
d) Remain stationary
Answer: a
Explanation: In complete inelastic motion, the e value is 0. Hence there is not relative motion between he two bodies. Therefore, the two bodies move together as a single body.