Two markers M1 and M2 are set up a vertical distance h apart. A steel ball is released at time zero from a point a distance x above M1.

Question 12

Two markers M1 and M2 are set up a vertical distance h apart.

A steel ball is released at time zero from a point a distance x above M1. The ball reaches M1 at time t1 and reaches M2 at time t2. The acceleration of the ball is constant.

Which expression gives the acceleration of the ball?

Reference: Past Exam Paper – November 2002 Paper 1 Q9 & June 2012 Paper 12 Q8

Solution:

Answer: D.

Consider the motion of the steel ball from the point of release to position M₁.

Initially, the ball is at rest. So, at time = 0, velocity u = 0 m s^-1.

Distance travelled to reach position M₁ is x.

s = ut + ½at²

x = 0(t₁) + ½ a t₁²       

x = ½ a t₁²                   ------------------ (1)

Now, consider the motion from the point of release to position M₂.

Initial velocity, u = 0

Distance travelled = x + h

Time taken = t₂

s = ut + ½ a t²

x + h = 0(t₂) + ½ a t₂²

x + h = ½ a t₂²             ------------------ (2)

We want to find the acceleration a and we do not need to find x specifically. So, we can eliminate x from the equations. 

Take (2) – (1),

x + h – x = ½at₂² – ½at₁²

h = ½ a (t₂² – t₁²)

a = 2h / (t₂² – t₁²)