A ball with a mass of  5 kg is rolling at 1 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Jun 28, 2018

The post collisions velocities are $= 011 m {s}^{-} 1$ and $= 1.411 m {s}^{-} 1$

Explanation:

Since the collision is elastic, we have conservation of momentum

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

And conservation of kinetic energy

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

The mass the first ball is ${m}_{1} = 5 k g$

The velocity of the first ball before the collision is ${u}_{1} = 1 m {s}^{-} 1$

The mass of the second ball is ${m}_{2} = 4 k g$

The velocity of the second ball before the collision is ${u}_{2} = 0 m {s}^{-} 1$

The velocity of the first ball after the collision is $= {v}_{1} m {s}^{-} 1$

The velocity of the second ball after the collision is $= {v}_{2} m {s}^{-} 1$

Therefore,

$5 \cdot 1 + 4 \cdot 0 = 5 \cdot {v}_{1} + 4 \cdot {v}_{2}$

$4 {v}_{2} + 5 {v}_{1} = 5$....................................$\left(1\right)$

$\frac{1}{2} \cdot 5 \cdot {1}^{2} + \frac{1}{2} \cdot 4 \cdot 0 = \frac{1}{2} \cdot 5 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$

$4 {v}_{2}^{2} + 5 {v}_{1}^{2} = 5$..............................$\left(2\right)$

Solving for ${v}_{1}$ and ${v}_{2}$ in equations $\left(1\right)$ and $\left(2\right)$

From equation $\left(1\right)$, ${v}_{2} = \frac{5 \left(1 - {v}_{1}\right)}{4}$

Substituting in equation $\left(2\right)$

$5 {v}_{1}^{2} + 4 \cdot {\left(\frac{5 \left(1 - {v}_{1}\right)}{4}\right)}^{2} = 5$

$4 {v}_{1}^{2} + 5 + 5 {v}_{1}^{2} - 10 {v}_{1} = 4$

$9 {v}_{1}^{2} - 10 {v}_{1} + 1 = 0$

${v}_{1} = \frac{10 \pm \sqrt{\left({\left(- 10\right)}^{2}\right) - 4 \left(9\right) \left(1\right)}}{18} = \frac{10 \pm 8}{18}$

Therefore,

${v}_{1} = 1 m {s}^{-} 1$ or ${v}_{1} = \frac{1}{9} = 0.11 m {s}^{-} 1$

${v}_{2} = 0 m {s}^{-} 1$ or ${v}_{2} = \frac{10}{9} = 1.11 m {s}^{-} 1$

The first solutions are the initial conditions.

The post collisions velocities are $= 011 m {s}^{-} 1$ and $= 1.411 m {s}^{-} 1$