Question

Three identical balls ball 1,ball 2 and ball 3 are placed on smooth floor on a straight line at the separation of 10m between each two balls.Ball 1 is given 10 m/s towards ball 2.Collision between ball 1 and ball 2 is inelastic with coefficient of restitution 0.5 but collision between ball 2 and ball 3 is perfectly elastic.The time interval between two consecutive collisions between ball 1 and ball 2 is _____________ seconds​.

Answer 1

• The time interval between two consecutive collisions between ball 1 and ball 2 is 4 seconds​.

There are three identical balls present in a smooth floor as shown in figure and having a separation of 10 m between each two balls.

Given-

Initial velocity of ball 1 = 10 m/sec

Coefficient of restitution = 0.5

There is no net external force working in the system. That's why we can apply the momentum conservation.

Balls are identical so they have a mass of m each .

Let velocity of 1 ball and 2 ball after collision be v₁ and v₂. So,

m × 10 = mv₁ + mv₂

10 = v₁ + v₂--------------(i)

We know that coefficient of restitution is minus of velocity of separation by velocity of approach.

e = -  (v₂ - v₁) / (0 - 10) = 0.5

v₂ - v₁ = 5------------(ii)

From equation (i) and (ii) we get-

v₂ = 7.5 m/sec

v₁ = 2.5 m/sec

Ball 2 after moving 10 m collides with ball 3 elastically (as shown in attached figure) and stops. But ball 1 moves towards ball 2. So, the time taken between two consecutive collisions is

T= 10 m/ 2.5 m/sec = 4 sec

Regards