Two balls X and Y are supported by long strings, as shown in Fig. 3.1. The balls are each pulled back and pushed towards each other.

Question 22

Two balls X and Y are supported by long strings, as shown in Fig. 3.1.

Fig. 3.1

The balls are each pulled back and pushed towards each other. When the balls collide at the position shown in Fig. 3.1, the strings are vertical. The balls rebound in opposite directions.

Fig. 3.2 shows data for X and Y during this collision.

Fig. 3.2

The positive direction is horizontal and to the right.

(a) Use the conservation of linear momentum to determine the mass M of Y. [3]

(b) State and explain whether the collision is elastic. [1]

(c) Use Newton’s second and third laws to explain why the magnitude of the change in momentum of each ball is the same. [3]

Reference: Past Exam Paper – June 2015 Paper 21 Q3

Solution:

(a)

{Momentum before collision = momentum after collision}

(4.5 × 50) + (– 2.8 × M) = (–1.8 × 50) + (1.4 × M)

(M = ) 75 g

(b) The total initial kinetic energy/KE is not equal to the total final kinetic energy/KE, so not elastic

(c) From Newton’s 3^rd law, the force on X is equal and opposite to the force on Y.

From Newton’s 2^nd law, the force equals/is proportional to rate of change of momentum (F = Δp/Δt)

Since the time of collision is the same for both balls, the change in momentum is also the same.

{Newton’s 2^nd law is about the ‘rate of change of momentum’ while the questions ask about the ‘change in momentum’. So, we need to make reference to the time of collision being the same to be able to compare the magnitude of change in momentum.}