Answer to Question #263468 in Mechanics | Relativity for tisjea

Question

A 0.1 kg ball moving at a constant velocity of 2 m/s strikes a 0.4 kg ball that is at rest. After the collision, the first ball rebounds straight back at 1.2 m/s. Calculate the final velocity of the second ball

Answer 1

The free-body diagram of the given problem is as follows:

The expression for the initial momentum of the ball is, P_i =m₁u₁ +m₂u₂

Substituting the given values in the above expression, we get

P_i = 0.1kg x 2 m/s + 0.4kg x 0 m/s

=0.2 kg m/s

The expression for the final momentum of the ball is,

P_f =m₁v₁ +m₂v₂

After the collision, a 0.1 kg ball is rebounding with 1.2 m/s in opposite direction with respect to the initial direction of velocity. Take the initial velocity direction or forward direction as a positive direction, then take the backward direction as negative.

Substituting the given values in the above expression, we get

P_f = 0.1kg x -1.2 m/s + 0.4 kg x v₂

₌ - 0.12kg m/s + 0.4 v₂

The principle of conservation of momentum states that the initial linear momentum of the system is equal to the final linear momentum of the system.

P_f=P_i

Substituting the given values in the above expression, we get

Thus, the final velocity of the second ball is 0.8 m/s.