Conservation of Momentum

Question

A red ball with a mass of 1 kg sits at rest on the gym floor. You roll a blue ball, with a mass of 100 kg, into the red ball. This blue ball is moving at 0.5 m/s before the collision. After the collision, the blue ball stops moving, and the red ball begins rolling. At what velocity would the red ball be rolling?

Ignore friction and assume it has completely elasticity.

This is being used for studying purposes only, im not completely sure if i got this correct so i'd like to compare my answer. If you can provide the step by step of how to do this also i'd appreciate it.

Answer 1

Conservation of Momentum states that all initial momenta (plural form of momentum) in the system equal all final momenta in the system after a collision has completed.

Mathematically stated:

∑p_i = ∑p_f

The summations are present because you don't pick and choose which momenta to consider; all momenta in both states of the collision must be considered or the math will make you think that the situation does not conserve momentum because the sides will be unequal.

Let's call the initial red ball momentum p_{R, i} and the initial blue ball momentum p_{B, i}

p_{R, i} = m_R * v_{R, i} = ( 1 kg ) * ( 0 m/s ) p_{R, i} = 0 kg * m/s

p_{B, i} = m_B * v_{B, i} = ( 100 kg ) * ( 0.5 m/s) p_{B, i} = 50 kg * m/s

∑p_i = p_{R, i} + p_{B, i} = ( 0 + 50 ) * kg * m/s ∑p_i = 50 kg * m/s

The problem prompt says the blue ball comes to rest, so v_{B, f} = 0 m/s.

The final red ball momentum is p_{R, f} and the final blue ball momentum is p_{B, f}.

p_{R, f} = m_R * v_{R, f} = (1 kg ) * ( v_{R, f} ) p_{R, f} = v_{R, f} * kg

p_{B, f} = m_B * v_{B, f} = ( 100 kg ) * ( 0 m/s ) p_{B, f} = 0 kg * m/s

∑p_f = p_{R, f} + p_{B, f} = 0 + v_{R, f} * kg ∑p_f = v_{R, f} * kg

∑p_i = ∑p_f 50 kg * m/s = v_{R, f} * kg Thus: v_{R, f} = 50 m/s after cancelling the kg units on both sides.

I hope this helps!