# A ball with a mass of  5 kg is rolling at 9 m/s and elastically collides with a resting ball with a mass of 2 kg. What are the post-collision velocities of the balls?

Apr 8, 2017

The velocity of the first ball is $= 3.86 m {s}^{-} 1$
The velocity of the second ball is $= 12.86 m {s}^{-} 1$

#### 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

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

${m}_{1} = 5 k g$

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

${u}_{1} = 9 m {s}^{-} 1$

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

Therefore,

${v}_{1} = \frac{3}{7} \cdot 9 + \frac{4}{7} \cdot \left(0\right) = \frac{27}{7} = 3.86 m {s}^{-} 1$

${v}_{2} = \frac{10}{7} \cdot 9 - \frac{3}{7} \cdot \left(0\right) = \frac{90}{7} = 12.86 m {s}^{-} 1$

Verificaition

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = 5 \cdot 9 + 2 \cdot 0 = 45$

${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = 5 \cdot 3.86 + 2 \cdot 12.86 = 45$