A ball with a mass of  6 kg is rolling at 18 m/s and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

Aug 5, 2017

${v}_{1} = - 3.6 \frac{m}{s}$ and ${v}_{2} = 14.4 \frac{m}{s}$

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 ${m}_{1} = 6 k g , {v}_{1 o} = 18 \frac{m}{s} , {m}_{2} = 9 k g ,$ and ${v}_{2 o} = 0$

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

${v}_{1} = \left(\frac{6 k g - 9 k g}{6 k g + 9 k g}\right) \left(18 \frac{m}{s}\right) + 0$

$= - 3.6 \frac{m}{s}$

the negative value indicates that the ball is now moving in the opposite direction that it approached from

${v}_{2} = \left(\frac{2 \left(6 k g\right)}{6 k g + 9 k g}\right) \left(18 \frac{m}{s}\right) - 0$

$= 14.4 \frac{m}{s}$

the positive value indicates that the second ball moves off in the direction opposite of the first

You can verify these answers using momentum and energy conservation.