# 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 3 kg. What are the post-collision velocities of the balls?

Nov 20, 2016

In a perfectly elastic collision, ${v}_{1} = 6 \frac{m}{s}$ and ${v}_{2} = 24 \frac{m}{s}$.

#### Explanation:

In a perfectly elastic collision, momentum and mechanical energy are conserved.

As two objects collide, all of the kinetic energy they possess is transferred into elastic potential energy in their molecular bonds, and then all of that energy which is stored in the bonds is transformed back into the post-collision kinetic energy of the objects. Although energy is transformed into potential energy during the collision, the mechanical energy before and after the collision is purely kinetic energy.

We are given that ${m}_{1} = 6 k g$, ${v}_{1} = 18 \frac{m}{s}$, ${m}_{2} = 3 k g$, and ${v}_{2} = 0$.

You can determine the final velocities in terms of the given quantities using these equations(1):

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

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

Because the second ball is initially at rest, these simplify to:

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

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

Therefore, the post-collision velocity of the first ball is:

${v}_{1} = \left(\frac{6 k g - 3 k g}{6 k g + 3 k g}\right) \left(18 \frac{m}{s}\right)$

${v}_{1} = 6 \frac{m}{s}$

and the post collision velocity of the second ball is:

${v}_{2} = \left(\frac{2 \left(6 k g\right)}{6 k g + 3 k g}\right) 18 \frac{m}{s}$

${v}_{2} = 24 \frac{m}{s}$

1. Knight, R. D. In Physics for Scientists and Engineers ; Houston, A., Ed.; Pearson: Glenview, IL, 2013; pp 265–266.