NEET Physics: Laws of Motion, Work-Energy and Rotation · Chapter 2

Work, Energy and Power

What to remember

Work W = F d cos θ. For a variable force, work is the area under the force-displacement graph. The work-energy theorem says: net work = change in kinetic energy.
Kinetic energy = ½ m v² = p²/2m. Spring potential energy = ½ k x². Mechanical energy (kinetic + potential) is conserved when only conservative forces act.
Power P = W/t = F · v. Collisions always conserve momentum. Kinetic energy is conserved only in elastic collisions, and the coefficient of restitution e = (speed of separation)/(speed of approach).
In a vertical circle with a string, the minimum speed at the top is √(g r) and at the bottom is √(5 g r). The tension at the bottom exceeds the tension at the top by 6 m g.

Work done by a constant force and a variable force

Work done by a constant force F acting through displacement d is W = F d cos θ = F · d (dot product), where θ is the angle between F and d. Work is a scalar. Its SI unit is the joule (J); 1 J = 1 N m = 1 kg m² s⁻². Dimensions: [M L² T⁻²].

Angle θ	Work	Example
0°	Positive, maximum (F d)	Pushing a box along the floor
90°	Zero	Centripetal force, normal reaction on a horizontal floor, a coolie carrying a load along a level road
180°	Negative (−F d)	Friction, force of the brake

For a variable force along a straight line, W = ∫ F dx from x₁ to x₂. On a force-displacement graph this is the area under the curve. Area above the axis is positive work and area below the axis is negative work. For a force that increases uniformly from zero to F over a distance d, the work is ½ F d.

The work done by gravity when a body of mass m falls through height h is m g h. When the body rises through h, gravity does −m g h.

Kinetic energy and the work-energy theorem

Kinetic energy K = ½ m v². It is always positive and a scalar. With momentum p = m v, K = p²/(2m) and p = √(2 m K). Two bodies with equal momentum: the lighter body has the larger kinetic energy. Two bodies with equal kinetic energy: the heavier body has the larger momentum.

Work-energy theorem: the work done by the net force on a body equals the change in its kinetic energy. W(net) = ½ m v² − ½ m u². The theorem holds for constant and variable forces and for conservative and non-conservative forces.

Worked example 1. A 2 kg body at rest is acted on by a net force of 10 N through 5 m. Work = 10 × 5 = 50 J. So ½ × 2 × v² = 50, giving v² = 50 and v = √50 ≈ 7.07 m s⁻¹. Check by kinematics: a = 5 m s⁻² and v² = 2 a s = 2 × 5 × 5 = 50.

A brake problem: a vehicle of mass m moving at speed v stops after a distance s under a constant retarding force F. Then F s = ½ m v², so the stopping distance s = m v²/(2F). The stopping distance is proportional to v². If the speed doubles, the distance becomes four times.

Potential energy, conservative forces and the spring

A force is conservative if the work done by it between two points does not depend on the path, or if the work done in a closed path is zero. Gravity, the spring force and the electrostatic force are conservative. Friction and air drag are non-conservative: the work done depends on the path and mechanical energy is lost as heat.

For a conservative force, the potential energy U is defined by F = − dU/dx.

Gravitational potential energy near the Earth: U = m g h (height h measured from a chosen reference level).
Spring potential energy: U = ½ k x², where k is the force constant (N m⁻¹) and x is the stretch or compression from the natural length. The spring force is F = − k x (Hooke's law). The work done in stretching a spring from 0 to x is ½ k x².

Spring energy stored is the same for stretch and compression of equal size. To stretch a spring from x₁ to x₂, the work is ½ k (x₂² − x₁²). A spring stretched to double its extension stores four times the energy.

Conservation of mechanical energy

If only conservative forces do work, the total mechanical energy K + U stays constant. For a freely falling body dropped from height h: at the top K = 0 and U = m g h; just before the ground K = ½ m v² = m g h, so v = √(2 g h). At any height y, ½ m v² + m g y = m g h.

Worked example 2. A block of mass 0.5 kg moving at 4 m s⁻¹ on a smooth horizontal surface hits a spring of force constant 200 N m⁻¹. The maximum compression x is found from ½ m v² = ½ k x². So 0.5 × 16 = 200 x², giving x² = 0.04 and x = 0.2 m.

When friction acts, the energy lost as heat equals the work done against friction. The total energy (including heat) is always conserved.

Power

Power is the rate of doing work. Average power P = W/t. Instantaneous power P = dW/dt = F · v = F v cos θ. SI unit: watt (W), 1 W = 1 J s⁻¹. Dimensions: [M L² T⁻³]. 1 horse power (hp) = 746 W.

Energy units: 1 kilowatt hour (kWh) = 1000 W × 3600 s = 3.6 × 10⁶ J. The kWh is a unit of energy, not of power. 1 electron volt = 1.6 × 10⁻¹⁹ J.

A machine lifting a load: power = m g v for lifting at constant speed v. A vehicle moving at constant speed on a level road against a resisting force F has engine power P = F v.

Worked example 3. A pump lifts 600 kg of water per minute from a well to a height of 20 m. Take g = 10 m s⁻². Work per minute = 600 × 10 × 20 = 1.2 × 10⁵ J. Power = 1.2 × 10⁵ / 60 = 2000 W = 2 kW.

Collisions

In every collision with no external force, total linear momentum is conserved. Kinetic energy is conserved only in elastic collisions.

Type	Momentum	Kinetic energy	Coefficient of restitution e
Perfectly elastic	Conserved	Conserved	1
Inelastic	Conserved	Partly lost	0 < e < 1
Perfectly inelastic	Conserved	Maximum loss (bodies move together after impact)	0

Coefficient of restitution: e = (relative speed of separation)/(relative speed of approach) = (v₂ − v₁)/(u₁ − u₂) along the line of impact.

Elastic collision in one dimension (masses m₁, m₂ with initial velocities u₁, u₂):

v₁ = [(m₁ − m₂) u₁ + 2 m₂ u₂] / (m₁ + m₂)
v₂ = [(m₂ − m₁) u₂ + 2 m₁ u₁] / (m₁ + m₂)

Special cases of an elastic collision when m₂ is at rest (u₂ = 0):

Equal masses (m₁ = m₂): the bodies exchange velocities; v₁ = 0 and v₂ = u₁.
A heavy body hits a light body at rest (m₁ ≫ m₂): the heavy body continues almost unchanged and the light body moves with about 2 u₁.
A light body hits a heavy body at rest (m₁ ≪ m₂): the light body rebounds with almost the same speed, v₁ ≈ −u₁, and the heavy body hardly moves.

Perfectly inelastic collision: the bodies stick together. Common velocity v = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). When m₂ is at rest, the fraction of kinetic energy lost is m₂/(m₁ + m₂).

A ball dropped from height h₁ that rebounds to height h₂ has e = √(h₂/h₁). The speed after the n-th bounce is eⁿ times the speed of impact.

Collisions in two dimensions, in brief: momentum is conserved separately along x and along y. For an elastic collision of equal masses with one at rest (not head-on), the two bodies move off at right angles to each other.

Motion in a vertical circle

A body of mass m is whirled in a vertical circle of radius r on a string. Let v be the speed at a point and T the tension.

At the lowest point: T − m g = m v²/r, so T(bottom) = m g + m v²/r.
At the highest point: T + m g = m v²/r, so T(top) = m v²/r − m g.
At the level of the centre (horizontal position): T = m v²/r, and the weight is tangential.

To complete the circle the string must stay taut at the top, so T(top) ≥ 0. The minimum speed at the top is v(top) = √(g r). By energy conservation between the bottom and the top: ½ m v(bottom)² = ½ m v(top)² + m g (2r). So v(bottom, min) = √(5 g r).

At minimum speed, T(top) = 0 and T(bottom) = m g + m (5 g r)/r = 6 m g. In general T(bottom) − T(top) = 6 m g. For a body at the end of a rigid rod (or inside a tube), the minimum speed at the top can be zero and at the bottom √(4 g r).

A body on the inside of a loop track (like a cyclist in a loop) must have the normal reaction ≥ 0 at the top, so the condition is the same: v(top) ≥ √(g r). At a point where the radius makes angle θ with the vertical (measured from the lowest point), v² = u² − 2 g r (1 − cos θ) and T = m v²/r + m g cos θ.

Formula summary

Quantity	Formula	Unit
Work (constant force)	W = F d cos θ	J
Work (variable force)	W = ∫ F dx	J
Kinetic energy	K = ½ m v² = p²/2m	J
Work-energy theorem	W(net) = ΔK	J
Spring force and energy	F = − k x; U = ½ k x²	N, J
Power	P = W/t = F v	W
Elastic collision (u₂ = 0)	v₁ = (m₁ − m₂) u₁/(m₁ + m₂); v₂ = 2 m₁ u₁/(m₁ + m₂)	m s⁻¹
Coefficient of restitution	e = (v₂ − v₁)/(u₁ − u₂)	none
Perfectly inelastic collision	v = (m₁ u₁ + m₂ u₂)/(m₁ + m₂)	m s⁻¹
Vertical circle	v(top) ≥ √(g r); v(bottom) ≥ √(5 g r)	m s⁻¹

Constants used: g = 9.8 m s⁻² (taken as 10 m s⁻² in rounded problems when stated).

NEET traps

Work is a scalar, but it can be positive, negative or zero. Do not confuse the sign of work with direction.
A force perpendicular to the displacement does no work. A person holding a load at rest does no work in the physical sense.
Kinetic energy is never negative, but potential energy can be negative depending on the reference level.
The work-energy theorem uses the net work of all forces, including friction. Do not use only the work of one force.
The kWh is a unit of energy, not of power. 1 kWh = 3.6 × 10⁶ J.
Momentum is conserved in every collision with no external force. Kinetic energy is conserved only in elastic collisions.
In a perfectly inelastic collision the kinetic energy loss is maximum, but not all of the kinetic energy is lost unless the total momentum is zero.
In a head-on elastic collision of equal masses the bodies exchange velocities. The body at rest does not always get the full speed in a non-head-on collision.
Minimum speed at the top of a vertical circle is √(g r), but at the bottom it is √(5 g r), not √(g r).
Equal momentum does not mean equal kinetic energy. For equal momentum, K ∝ 1/m.

One-liners

1 joule = 1 newton metre = 1 kg m² s⁻².
The work done by a spring while it is compressed by x is − ½ k x² (the work done by us is + ½ k x²).
The area under a force-displacement graph is work.
Work done by a conservative force over a closed path is zero.
The work done by the net force is equal to the change in kinetic energy.
The work done by the centripetal force is always zero.
1 kWh = 3.6 × 10⁶ J.
1 horse power = 746 W.
Power = force × velocity.
For a coefficient of restitution of 1 the collision is perfectly elastic; for 0 it is perfectly inelastic.
In a head-on elastic collision of equal masses the velocities are exchanged.
A light body hitting a heavy body at rest in an elastic collision bounces back with nearly the same speed.
K = p²/2m links kinetic energy and momentum.
At the top of a vertical circle the minimum speed is √(g r) and at the bottom it is √(5 g r).
The tension at the bottom of a vertical circle is larger than at the top by 6 m g.

Practice questions

A constant force of 10 N acts on a body at 60° to the direction of its displacement of 4 m. The work done by the force is
1. 10 J
2. 20√3 J
3. 40 J
4. 20 J
Answer
D. 20 J
W = F d cos 60° = 10 × 4 × 0.5 = 20 J.
The kinetic energy of a 2 kg body moving at 6 m s⁻¹ is
1. 72 J
2. 18 J
3. 36 J
4. 12 J
Answer
C. 36 J
K = ½ m v² = ½ × 2 × 36 = 36 J. Check with p = 12: p²/2m = 144/4 = 36.
A force F = 3x newton acts on a body along the x-axis, where x is in metres. The work done in moving the body from x = 0 to x = 4 m is
1. 48 J
2. 24 J
3. 12 J
4. 36 J
Answer
B. 24 J
W = ∫3x dx = 1.5 x² = 1.5 × 16 = 24 J. Check by area of the triangle: ½ × 4 × 12 = 24 J.
A spring of force constant 400 N m⁻¹ is stretched by 5 cm. The potential energy stored in it is
1. 0.5 J
2. 100 J
3. 1.0 J
4. 10 J
Answer
A. 0.5 J
U = ½ k x² = ½ × 400 × (0.05)² = 0.5 J. Check: average force 10 N × 0.05 m = 0.5 J.
A motor lifts a 100 kg load through 10 m in 20 s at constant speed. Taking g = 10 m s⁻², the power delivered to the load is
1. 5000 W
2. 500 W
3. 200 W
4. 1000 W
Answer
B. 500 W
Work = 100 × 10 × 10 = 10⁴ J; P = 10⁴/20 = 500 W.
An electric heater consumes 2 kilowatt hour of energy. This is equal to
1. 7.2 × 10⁶ J
2. 2 × 10⁶ J
3. 7.2 × 10³ J
4. 3.6 × 10⁶ J
Answer
A. 7.2 × 10⁶ J
1 kWh = 3.6 × 10⁶ J, so 2 kWh = 7.2 × 10⁶ J.
A 2 kg ball moving at 6 m s⁻¹ collides head-on and elastically with a 4 kg ball at rest. The speed of the 4 kg ball after the collision is
1. 2 m s⁻¹
2. 6 m s⁻¹
3. 3 m s⁻¹
4. 4 m s⁻¹
Answer
D. 4 m s⁻¹
v₂ = 2 m₁u₁/(m₁ + m₂) = 2 × 2 × 6/6 = 4 m s⁻¹. Check by momentum: 12 = 2(−2) + 4(4) = 12.
A body of mass 3 kg moving at 4 m s⁻¹ hits a stationary body of mass 1 kg and the two stick together. Their common speed is
1. 1 m s⁻¹
2. 12 m s⁻¹
3. 3 m s⁻¹
4. 4 m s⁻¹
Answer
C. 3 m s⁻¹
3 × 4 = (3 + 1)v gives v = 3 m s⁻¹.
A small body is whirled in a vertical circle of radius 0.9 m on a light string. Taking g = 10 m s⁻², the minimum speed at the highest point for the string to stay taut is
1. 9 m s⁻¹
2. 4.5 m s⁻¹
3. 3 m s⁻¹
4. 3√5 m s⁻¹
Answer
C. 3 m s⁻¹
v = √(g r) = √9 = 3 m s⁻¹.
A body is whirled in a vertical circle of radius 0.5 m on a light string. Taking g = 10 m s⁻², the minimum speed at the lowest point to complete the circle is
1. √5 m s⁻¹
2. 2.5 m s⁻¹
3. 25 m s⁻¹
4. 5 m s⁻¹
Answer
D. 5 m s⁻¹
v = √(5 g r) = √25 = 5 m s⁻¹. Check by energy: ½v² = ½(√5)² + 10 × 1 gives v² = 5 + 20 = 25.
A stone of mass 0.5 kg moves in a vertical circle on a string. The tension at the lowest point exceeds the tension at the highest point by (g = 10 m s⁻²)
1. 15 N
2. 30 N
3. 25 N
4. 10 N
Answer
B. 30 N
T(bottom) − T(top) = 6 m g = 6 × 0.5 × 10 = 30 N.
A ball dropped from a height of 1.6 m rebounds to a height of 0.9 m. The coefficient of restitution is
1. 0.75
2. 0.25
3. 0.5
4. 0.5625
Answer
A. 0.75
e = √(h₂/h₁) = √(0.9/1.6) = 0.75. Check: speeds before and after are √(2g × 1.6) and √(2g × 0.9), ratio 3/4.
A car moving at 10 m s⁻¹ is stopped by brakes in 20 m. With the same braking force, the stopping distance from 20 m s⁻¹ is
1. 160 m
2. 20 m
3. 80 m
4. 40 m
Answer
C. 80 m
s ∝ v², so s = 20 × 4 = 80 m. Check: F s = ½ m v² with the same F.
A car engine of power 20 kW drives the car at a constant speed of 20 m s⁻¹. The resisting force on the car is
1. 1000 N
2. 400 N
3. 10 N
4. 100 N
Answer
A. 1000 N
P = F v gives F = 20000/20 = 1000 N.
A block of mass 5 kg slides 2 m on a horizontal surface with μk = 0.2. Taking g = 10 m s⁻², the work done by the friction force is
1. +20 J
2. −200 J
3. −10 J
4. −20 J
Answer
D. −20 J
f = 0.2 × 50 = 10 N; W = −f d = −10 × 2 = −20 J.
Which of the following forces is non-conservative?
1. Gravitational force
2. Friction
3. Spring force
4. Electrostatic force
Answer
B. Friction
The work done by friction depends on the path.
A 2 kg block slides from rest down a rough incline of vertical height 5 m and reaches the bottom at 6 m s⁻¹. Taking g = 10 m s⁻², the work done against friction is
1. 136 J
2. 36 J
3. 64 J
4. 100 J
Answer
C. 64 J
Loss = m g h − ½ m v² = 100 − 36 = 64 J.
A moving ball of mass m collides elastically, but not head-on, with an identical ball at rest. The angle between the directions of the two balls after the collision is
1. 180°
2. 90°
3. 0°
4. 45°
Answer
B. 90°
Momentum and kinetic energy conservation give v₁ · v₂ = 0, so the velocities are perpendicular.
A bullet of mass 0.02 kg moving at 300 m s⁻¹ embeds in a 1.98 kg block at rest. The percentage of the initial kinetic energy lost is
1. 90%
2. 1%
3. 50%
4. 99%
Answer
D. 99%
Fraction lost = m₂/(m₁ + m₂) = 1.98/2 = 0.99. Check: v = 6/2 = 3 m s⁻¹, final K = 9 J, initial K = 900 J.
A crane lifts a 500 kg load at a constant speed of 2 m s⁻¹. Taking g = 10 m s⁻², the power it delivers is
1. 10 kW
2. 5 kW
3. 20 kW
4. 1 kW
Answer
A. 10 kW
P = m g v = 500 × 10 × 2 = 10⁴ W.
A spring of force constant 200 N m⁻¹ is compressed by 0.1 m and releases a 20 g ball on a smooth horizontal surface. The speed of the ball is
1. 100 m s⁻¹
2. 5 m s⁻¹
3. 20 m s⁻¹
4. 10 m s⁻¹
Answer
D. 10 m s⁻¹
½ k x² = 1 J = ½ (0.02) v², so v² = 100 and v = 10 m s⁻¹.
A 0.2 kg stone is whirled in a vertical circle of radius 1 m. Its speed at the lowest point is 10 m s⁻¹. Taking g = 10 m s⁻², the tension in the string at the lowest point is
1. 18 N
2. 22 N
3. 20 N
4. 2 N
Answer
B. 22 N
T = m g + m v²/r = 2 + 0.2 × 100 = 22 N.
Two bodies of masses 1 kg and 4 kg have equal momentum. The ratio of the kinetic energy of the 1 kg body to that of the 4 kg body is
1. 2 : 1
2. 1 : 2
3. 4 : 1
4. 1 : 4
Answer
C. 4 : 1
K = p²/2m, so K ∝ 1/m and the ratio = 4/1.
Given below are two statements: Statement I: Of two bodies with equal momentum, the lighter body has the larger kinetic energy. Statement II: Kinetic energy is p²/2m.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
K = p²/2m is inversely proportional to mass for equal p.
Given below are two statements: Statement I: The kilowatt hour is a unit of power. Statement II: 1 kWh = 3.6 × 10⁶ J.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
kWh is a unit of energy.
Given below are two statements: Statement I: In an inelastic collision the kinetic energy is conserved. Statement II: In an inelastic collision the total momentum is not conserved.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
In an inelastic collision kinetic energy is lost but momentum is conserved.
Which of the following statements are correct? 1. The work done by a conservative force over a closed path is zero. 2. Friction is a conservative force. 3. The spring force is a conservative force.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Friction is non-conservative, so statement 2 is wrong.
Given below are two statements: Statement I: Potential energy can be negative. Statement II: Kinetic energy can be negative.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
K = ½ m v² is never negative; U depends on the chosen reference level.
Given below are two statements: Statement I: Gravity does zero work on a body moving along a horizontal path. Statement II: Gravity acts perpendicular to the horizontal displacement.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
W = m g d cos 90° = 0.
Which of the following statements are correct? 1. For a perfectly inelastic collision e = 0. 2. In a perfectly inelastic collision kinetic energy is conserved. 3. Total momentum is conserved in all collisions without external force.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 3 only
Kinetic energy is not conserved in a perfectly inelastic collision.
Given below are two statements: Statement I: A body at the lowest point of a vertical circle needs a speed of √(g r) to complete the circle. Statement II: The minimum speed at the lowest point is √(5 g r).
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
√(g r) is the minimum at the top; √(5 g r) at the bottom.
Given below are two statements: Statement I: Power equals the dot product of force and velocity. Statement II: Power is a vector quantity.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
P = F · v is a scalar.
Given below are two statements: Statement I: The energy stored in a spring is different for a stretch x and for a compression x. Statement II: The energy stored in a spring depends on x².
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
U = ½ k x² has the same value for +x and −x.
Assertion (A): A porter carrying luggage on his head along a level platform does no work on the luggage. Reason (R): The force applied by the porter is perpendicular to the displacement.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The porter's force is vertical and the displacement is horizontal, so W = 0.
Assertion (A): All the kinetic energy is lost in a perfectly inelastic collision. Reason (R): In a perfectly inelastic collision the bodies stick together after the collision.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Sticking together gives maximum loss, but the common velocity is generally not zero, so not all is lost. R is true.
Assertion (A): A force perpendicular to the displacement does no work. Reason (R): The centripetal force does no work in uniform circular motion.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true. R is an example, not the reason.
Assertion (A): A ball dropped from a height rebounds to a lower height from the floor. Reason (R): The collision of the ball with the floor is perfectly elastic.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
The collision is inelastic, so R is false.
Assertion (A): When a body of mass m hits an identical body at rest elastically and head-on, the first body stops. Reason (R): In an elastic collision the total momentum is conserved.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the stopping needs conservation of kinetic energy as well, so R alone is not the explanation.
Assertion (A): The tension in a string whirled in a vertical circle is the same at all points. Reason (R): The speed of the body changes as it moves along the circle.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
The tension changes with position; the speed does change.
Match List I with List II. List I: a. Work b. Power c. Spring constant d. Coefficient of restitution. List II: i. dimensionless ii. [M L² T⁻³] iii. [M T⁻²] iv. [M L² T⁻²]
1. a-iv, b-ii, c-iii, d-i
2. a-iii, b-ii, c-iv, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-ii, b-iv, c-iii, d-i
Answer
A. a-iv, b-ii, c-iii, d-i
Work [M L² T⁻²]; power [M L² T⁻³]; k = F/x [M T⁻²]; e is a ratio.
Match List I with List II. List I: a. Perfectly elastic collision b. Perfectly inelastic collision c. Inelastic collision d. Heavy body hitting a light body at rest, elastically. List II: i. e = 0 ii. 0 < e < 1 iii. e = 1 iv. light body moves at about twice the speed of the heavy body
1. a-iii, b-i, c-ii, d-iv
2. a-i, b-iii, c-ii, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-iii, b-i, c-iv, d-ii
Answer
A. a-iii, b-i, c-ii, d-iv
e values are 1, 0 and between 0 and 1; v₂ ≈ 2u₁ for m₁ ≫ m₂.
Match List I with List II for a body on a string in a vertical circle of radius r. List I: a. Minimum speed at the top b. Minimum speed at the bottom c. Tension at the top at minimum speed d. Tension at the bottom at minimum speed. List II: i. 6 m g ii. √(5 g r) iii. zero iv. √(g r)
1. a-ii, b-iv, c-iii, d-i
2. a-iv, b-ii, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-iii, c-ii, d-i
Answer
B. a-iv, b-ii, c-iii, d-i
At minimum speed T(top) = 0, and T(bottom) = m g + 5 m g = 6 m g.
Match List I with List II. List I: a. Work of a 20 N force along a 3 m displacement b. Kinetic energy of a 1 kg body at 4 m s⁻¹ c. Energy of a spring with k = 100 N m⁻¹ stretched by 0.2 m d. Energy used by a 100 W lamp in 5 s. List II: i. 2 J ii. 500 J iii. 60 J iv. 8 J
1. a-i, b-iv, c-iii, d-ii
2. a-iv, b-iii, c-i, d-ii
3. a-iii, b-iv, c-i, d-ii
4. a-iii, b-iv, c-ii, d-i
Answer
C. a-iii, b-iv, c-i, d-ii
60 J; 8 J; ½ × 100 × 0.04 = 2 J; 100 × 5 = 500 J.
Match List I with List II for the final velocity v₁ of the first body (mass m₁, speed u) that hits a second body (mass m₂) at rest. List I: a. Elastic, m₁ = m₂ b. Elastic, m₁ ≫ m₂ c. Elastic, m₁ ≪ m₂ d. Perfectly inelastic, m₁ = m₂. List II: i. −u ii. 0 iii. u iv. u/2
1. a-ii, b-iii, c-i, d-iv
2. a-ii, b-iii, c-iv, d-i
3. a-i, b-iii, c-ii, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
A. a-ii, b-iii, c-i, d-iv
Exchange of velocities gives 0; heavy body keeps about u; light body returns with about −u; sticking gives u/2.
Match List I with List II. List I: a. Friction on a sliding block b. Gravity on a falling body c. Centripetal force in uniform circular motion d. Work done by an external agent in stretching a spring from natural length by x. List II: i. zero ii. positive iii. negative iv. + ½ k x²
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-i, b-ii, c-iii, d-iv
Answer
B. a-iii, b-ii, c-i, d-iv
Friction opposes motion (negative); gravity acts along the fall (positive); centripetal force does zero work; the agent does + ½ k x².

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1.The blades of a windmill rotate because of which of the following?
2.The formula to find the power is-
3.The elastic potential energy of a strained body is:
4.The energy possessed by a body due to its position or configuration is called ________.
5.The energy possessed by a ball of mass 5 kg when it is raised to a height of 5 m above the ground is (take g = 10 m/s 2):
6.The energy possessed by an object by virtue of its motion is known as ________.
7.The energy possessed by a moving object due to its motion is called:
8.The energy that is derived from the use of radioactive isotopes is termed as _________.
9.The SI unit of power is______.
10.The sum of _______ is mechanical energy.

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