# A ball with a mass of  6 kg is rolling at 16 m/s and elastically collides with a resting ball with a mass of  1 kg. What are the post-collision velocities of the balls?

Apr 10, 2018

#### Answer:

These momentum problems are the first in physics where you're responsible for managing a (relatively) large number of variables.

#### Explanation:

Recall,

${\nu}_{A} - {\nu}_{B} = {\nu}_{B} ' - {\nu}_{A} '$

Which is derived starting from,

${m}_{A} \left({\nu}_{A} - {\nu}_{A} '\right) = {m}_{b} \left({\nu}_{B} ' - {\nu}_{B}\right)$ $\left(1\right)$

${m}_{A} \left({\nu}_{A}^{2} - {\nu}_{A}^{' 2}\right) = {m}_{B} \left({\nu}_{B}^{' 2} - {\nu}_{B}^{2}\right)$ $\left(2\right)$

(a restatement of the law of conservation of momentum and energy, respectively).

Then, by factoring out (2), and dividing the expansion by (1).

From here, since ball B is at rest, it's easy. Consider,

${\nu}_{A} = {\nu}_{B} ' - {\nu}_{A} '$

$\implies {\nu}_{B} ' = {\nu}_{A} + {\nu}_{A} '$

Now, consider the conformity with the law of conservation of momentum, and substituting this variable in,

${\nu}_{A} = {\nu}_{A} ' + {\nu}_{A} + {\nu}_{A} '$

nu_A' = (0"m")/"s",

and by extension,

nu_A = nu_B' = (16"m")/"s"

From this calculation, we can understand that:

In perfectly elastic collisions, all of the energy or momentum is transferred perfectly from one body to the other, if one is at rest.