Question

A 3.0 kg hard steel ball collides head on with a 1.0 kg hard steel ball. The balls are moving at 2 m/s in opposite directions before they collide. Upon colliding, the 3.0 kg ball stops. What is the velocity of the 1.0 kg ball after the collision?

Answer 1

Step-by-step explanation:

The Law of Momentum Conservation applies and its equation is:

`[(m_1v_1)+(m_2v_2)]_b=[(m_1v_1)+(m_2v_2)]_a` and filling in and solving for v2:

`[(3.0*2.0)+(1.0*-2.0)]_b=[(3.0*0)+(1.0v_2)]_a` and

6.0 - 2.0 = 0 + 1.0v₂ and

4.0 = v₂ in the initial direction of the 3.0 kg ball (since the velocity of the 3.0 kg ball is positive and the velocity of the 1.0 kg ball is negative and our answer is positive).