A ball with a mass of  6 kg is rolling at 24 m/s and elastically collides with a resting ball with a mass of 3 kg. What are the post-collision velocities of the balls?

Jul 7, 2017

${v}_{1} = 8 \frac{m}{s}$ and ${v}_{2} = 32 \frac{m}{s}$

Explanation:

Momentum is conserved in all collisions, but an elastic collision is one in which both momentum and mechanical energy are conserved. For elastic collisions, we use these equations to determine unknown values:

${v}_{1} = \left(\frac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{1 o} + \left(\frac{2 {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{2 o}$

${v}_{2} = \left(\frac{2 {m}_{1}}{{m}_{1} + {m}_{2}}\right) {v}_{1 o} - \left(\frac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{2 o}$

We are given ${m}_{1} = 6 k g , {v}_{1 o} = 24 \frac{m}{s} , {m}_{2} = 3 k g ,$ and ${v}_{2 o} = 0$

We can use each equation to solve for ${v}_{1}$ and ${v}_{2}$.

${v}_{1} = \left(\frac{6 k g - 3 k g}{6 k g + 3 k g}\right) \left(24 \frac{m}{s}\right) + 0$

$= 8 \frac{m}{s}$

${v}_{2} = \left(\frac{2 \left(6 k g\right)}{6 k g + 3 k g}\right) \left(24 \frac{m}{s}\right) - 0$

$= 32 \frac{m}{s}$

You can verify this answer using momentum and energy conservation.