Question

11. (a) State the principle of conservation of linear momentum.
(b) Explain the mode of action of a 'propelled rocket'.
(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.
(d)(i) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height. Would the balls hit the ground at the same time? (ii) Explain your answer.
(e) A ball of mass 0.10kg is projected horizontally onto a vertical wall with a speed of 17ms⁻¹. The ball makes contact with the wall for 0.15s and rebounds horizontally with a speed of 13ms⁻¹. Calculate the: (i) change in momentum of the ball; (ii) average force exerted on the ball during its collision with the wall.

Accepted Answer

(a) The principle of conservation of linear momentum states that in a closed system (with no external force acting), the total momentum before a collision (or interaction) is equal to the total momentum after the collision.

(b) A propelled rocket operates by burning fuel to expel hot gases at high speed out of the rear of the rocket. By Newton's third law (and conservation of momentum), the backward momentum of the expelled gases is balanced by an equal and opposite (forward) momentum gained by the rocket, propelling it forward.

(c) Reasons the ball's motion is not simple harmonic motion: (i) The restoring force on the ball is not proportional to its displacement from a fixed equilibrium/mean position. (ii) The motion of the ball does not oscillate about a fixed central point with the ball's acceleration always directed towards that point.

(d)(i) Yes, the two balls would hit the ground at the same time. (ii) This is because the vertical motion of both balls is independent of any horizontal motion; both start with zero vertical velocity and experience the same vertical acceleration due to gravity, so they take the same time to fall the same vertical height, regardless of their horizontal velocities.

(e)(i) Change in momentum = m(v−u), taking the direction towards the wall as positive: initial momentum = 0.10×17 = 1.7kgms⁻¹ (towards wall); final momentum = 0.10×(−13) = −1.3kgms⁻¹ (away from wall, i.e. rebounding). Change in momentum = final − initial = −1.3−1.7 = −3.0kgms⁻¹ (magnitude 3.0kgms⁻¹).
(ii) Average force = change in momentum/time = 3.0/0.15 = 20N.

WAEC Physics 2012 Theory — Question 11

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