# A ball with a mass of 3 kg is rolling at 3 m/s and elastically collides with a resting ball with a mass of 1 kg. What are the post-collision velocities of the balls?

Jan 2, 2016

Equations of conservation of energy and momentum.

${u}_{1} ' = 1.5 \frac{m}{s}$

${u}_{2} ' = 4.5 \frac{m}{s}$

#### Explanation:

As wikipedia suggests:

${u}_{1} ' = \frac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}} \cdot {u}_{1} + \frac{2 {m}_{2}}{{m}_{1} + {m}_{2}} \cdot {u}_{2} =$

$= \frac{3 - 1}{3 + 1} \cdot 3 + \frac{2 \cdot 1}{3 + 1} \cdot 0 =$

$= \frac{2}{4} \cdot 3 = 1.5 \frac{m}{s}$

${u}_{2} ' = \frac{{m}_{2} - {m}_{1}}{{m}_{1} + {m}_{2}} \cdot {u}_{2} + \frac{2 {m}_{1}}{{m}_{1} + {m}_{2}} \cdot {u}_{1} =$

$= \frac{1 - 3}{3 + 1} \cdot 0 + \frac{2 \cdot 3}{3 + 1} \cdot 3 =$

$= - \frac{2}{4} \cdot 0 + \frac{6}{4} \cdot 3 = 4.5 \frac{m}{s}$

[Equations' source]

Derivation

Conservation of momentum and energy state:

Momentum

${P}_{1} + {P}_{2} = {P}_{1} ' + {P}_{2} '$

Since momentum is equal to $P = m \cdot u$

${m}_{1} \cdot {u}_{1} + {m}_{2} \cdot {u}_{2} = {m}_{1} \cdot {u}_{1} ' + {m}_{2} \cdot {u}_{2} '$ - - - $\left(1\right)$

Energy

${E}_{1} + {E}_{2} = {E}_{1} ' + {E}_{2} '$

Since kinetic energy is equal to $E = \frac{1}{2} \cdot m \cdot {u}^{2}$

$\frac{1}{2} \cdot {m}_{1} \cdot {u}_{1}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {u}_{2}^{2} = \frac{1}{2} \cdot {m}_{1} \cdot {u}_{1}^{2} ' + \frac{1}{2} \cdot {m}_{2} \cdot {u}_{2}^{2} '$ - - - $\left(2\right)$

You can use $\left(1\right)$ and $\left(2\right)$ to prove the equations mentioned above. (I tried but kept getting two solutions, which is not right)