# A ball with a mass of  3 kg is rolling at 5 m/s and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

May 7, 2017

The velocity of the first ball is $= - 2.5 m {s}^{-} 1$
The velocity of the second ball is $= 2.5 m {s}^{-} 1$

#### Explanation:

In an elastic collision, we have conservation of momentum and conservation of kinetic energy

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

and

$\frac{1}{2} {m}_{1} {u}_{1}^{2} + \frac{1}{2} {m}_{2} {u}_{2}^{2} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2}$

Solving the above 2 equations for ${v}_{1}$ and ${v}_{2}$, we get

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

and

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

Taking the direction as positive ${\rightarrow}^{+}$

${m}_{1} = 3 k g$

${m}_{2} = 9 k g$

${u}_{1} = 5 m {s}^{-} 1$

${u}_{2} = 0 m {s}^{-} 1$

Therefore,

${v}_{1} = - \frac{6}{12} \cdot 5 + \frac{18}{12} \cdot \left(0\right) = \frac{27}{7} = - 2.5 m {s}^{-} 1$

${v}_{2} = \frac{6}{12} \cdot 5 - \frac{6}{12} \cdot \left(0\right) = \frac{90}{7} = 2.5 m {s}^{-} 1$

Verificaition

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = 3 \cdot 5 + 9 \cdot 0 = 15$

${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = - 3 \cdot 2.5 + 9 \cdot 2.5 = 6 \cdot 2.5 = 15$