A ball with a mass of 2 kg is rolling at 4 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

May 9, 2017

The velocity of the first ball is $= - 1.33 m {s}^{-} 1$
The velocity of the second ball is $= 2.67 m {s}^{-} 1$

Explanation:

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

The velocities before the collision are ${u}_{1}$ and ${u}_{2}$.

The velocities after the collision are ${v}_{1}$ and ${v}_{2}$.

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

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 ${\rightarrow}^{+}$

${m}_{1} = 2 k g$

${m}_{2} = 4 k g$

${u}_{1} = 4 m {s}^{-} 1$

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

Therefore,

${v}_{1} = - \frac{2}{6} \cdot 4 + \frac{8}{6} \cdot \left(0\right) = - \frac{4}{3} = - 1.33 m {s}^{-} 1$

${v}_{2} = \frac{4}{6} \cdot 4 - \frac{2}{6} \cdot \left(0\right) = \frac{8}{3} = 2.67 m {s}^{-} 1$

Verificaition

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = 2 \cdot 4 + 4 \cdot 0 = 8$

${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = - 2 \cdot \frac{4}{3} + 4 \cdot \frac{8}{3} = 8$