Work, Energy, Power and Collisions — Practice

Question 1

A force of 10 N pulls a body through 5 m at 60° to the displacement. The work done is:
(a) 25 J (b) 50 J (c) 43 J (d) 12.5 J

Accepted Answer

(a) 25 J.

W = Fd cosθ = 10 × 5 × cos60° = 10 × 5 × 0.5 = 25 J.

Question 2

A satellite moves in a circular orbit. The work done by the gravitational force on it over one complete revolution is:
(a) zero (b) positive (c) negative (d) equal to its kinetic energy

Accepted Answer

(a) zero.

In a circular orbit gravity is always perpendicular to the velocity (θ = 90°), so it does no work — which is why the orbital speed stays constant.

Question 3

A net work of 100 J is done on a 2 kg body initially at rest. Its final speed is:
(a) 10 m/s (b) 5 m/s (c) 50 m/s (d) 100 m/s

Accepted Answer

(a) 10 m/s.

W_net = ½mv² − 0, so 100 = ½ × 2 × v² = v², giving v = 10 m/s.

Question 4

The gravitational potential energy of a 2 kg mass at a height of 5 m is (g = 10 m/s²):
(a) 100 J (b) 10 J (c) 50 J (d) 20 J

Accepted Answer

(a) 100 J.

PE = mgh = 2 × 10 × 5 = 100 J.

Question 5

A spring stores energy E when stretched by x. If it is stretched by 2x instead, the stored energy is:
(a) 4E (b) 2E (c) E/2 (d) 8E

Accepted Answer

(a) 4E.

Spring energy = ½kx² ∝ x². Doubling the extension multiplies the stored energy by 2² = 4.

Question 6

A body is dropped from a height of 20 m. Its speed just before hitting the ground is (g = 10 m/s², ignore air resistance):
(a) 20 m/s (b) 10 m/s (c) 40 m/s (d) 14 m/s

Accepted Answer

(a) 20 m/s.

Energy conservation: ½mv² = mgh, so v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s — independent of mass.

Question 7

A force of 20 N moves a body at a constant 5 m/s in its own direction. The power delivered is:
(a) 100 W (b) 4 W (c) 25 W (d) 15 W

Accepted Answer

(a) 100 W.

P = Fv = 20 × 5 = 100 W.

Question 8

A pump raises 100 kg of water through 10 m in 20 s. Its power output is (g = 10 m/s²):
(a) 500 W (b) 1000 W (c) 250 W (d) 5000 W

Accepted Answer

(a) 500 W.

Work = mgh = 100 × 10 × 10 = 10 000 J; power = work/time = 10 000/20 = 500 W.

Question 9

If the momentum of a body is doubled (mass unchanged), its kinetic energy becomes:
(a) 4 times (b) 2 times (c) √2 times (d) unchanged

Accepted Answer

(a) 4 times.

KE = p²/2m ∝ p². Doubling the momentum multiplies the kinetic energy by 2² = 4.

Question 10

A ball moving at speed v collides head-on and elastically with an identical ball at rest. After the collision, the first ball:
(a) stops (b) continues at v (c) rebounds at v (d) moves at v/2

Accepted Answer

(a) stops.

For equal masses in a head-on elastic collision with one at rest, the velocities exchange: the incoming ball stops and the target moves off at v.

Question 11

A 2 kg body moving at 6 m/s strikes a stationary 4 kg body and they stick together. Their common speed is:
(a) 2 m/s (b) 3 m/s (c) 6 m/s (d) 1 m/s

Accepted Answer

(a) 2 m/s.

Momentum conservation: 2 × 6 = (2 + 4)V, so V = 12/6 = 2 m/s. Kinetic energy drops from 36 J to 12 J — the rest is lost as heat and deformation.

Question 12

The coefficient of restitution for a perfectly elastic collision is:
(a) 1 (b) 0 (c) 0.5 (d) infinite

Accepted Answer

(a) 1.

For a perfectly elastic collision the relative speed of separation equals the relative speed of approach, so e = 1. A perfectly inelastic collision has e = 0.

Question 13

The minimum speed at the topmost point for a body to just complete a vertical circle of radius r on a string is:
(a) √(gr) (b) √(2gr) (c) √(5gr) (d) √(gr/2)

Accepted Answer

(a) √(gr).

At the top, gravity alone must supply the centripetal force: mg = mv²/r, so v_top = √(gr).

Question 14

For the same vertical circle of radius r, the minimum speed at the lowest point to just complete the loop is:
(a) √(5gr) (b) √(gr) (c) √(2gr) (d) √(3gr)

Accepted Answer

(a) √(5gr).

Energy conservation from top to bottom (height 2r): ½mv_bottom² = ½mv_top² + mg(2r). With v_top² = gr, v_bottom² = gr + 4gr = 5gr, so v_bottom = √(5gr).

Question 15

A coolie carries a 20 kg load on his head and walks 50 m on level ground. The work done against gravity is:
(a) zero (b) 10 000 J (c) 1000 J (d) 500 J

Accepted Answer

(a) zero.

Gravity acts vertically while the displacement is horizontal (θ = 90°), so W = Fd cos90° = 0. No work is done against gravity on a level walk, however tiring it feels.

Practice — Work, Energy, Power and Collisions

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