# A ball with a mass of  3  kg is rolling at 18  ms^-1 and elastically collides with a resting ball with a mass of 9  kg. What are the post-collision velocities of the balls?

Feb 24, 2016

In an elastic collision, both momentum and kinetic energy are conserved. If we call the $3$ $k g$ ball 1 and the $9$ $k g$ ball 2, their final velocities are ${v}_{1} = - 14.4$ and ${v}_{2} = 10.8$ $m {s}^{-} 1$.

#### Explanation:

Initial momentum:

$p = {m}_{1} {v}_{1} + {m}_{2} {v}_{2} = 3 \cdot 18 + 9 \cdot 0 = 54$ $k g m {s}^{-} 1$

Initial kinetic energy:

${E}_{k} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{1}{2} \cdot 3 \cdot {18}^{2} + \frac{1}{2} \cdot 9 \cdot {0}^{2} = 486$ $J$

Because this is an elastic collision, both will be conserved:

Final momentum:

$p = 54 = {m}_{1} {v}_{1} + {m}_{2} {v}_{2} = 3 {v}_{1} + 9 {v}_{2}$ - call this $\left(1\right)$

Final kinetic energy:

${E}_{k} = 486 = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{1}{2} \cdot 3 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 9 \cdot {v}_{2}^{2}$ - call this $\left(2\right)$

We now have two equations, $\left(1\right)$ and $\left(2\right)$, in two unknowns.

Let's double $\left(2\right)$, just for neatness:

$972 = 3 {v}_{1}^{2} + 9 {v}_{2}^{2}$

We can find a value for ${v}_{1}$ by rearranging $\left(1\right)$ and then substitute that into this revised $\left(2\right)$ so that we are working with only one variable:

$54 = 3 {v}_{1} + 9 {v}_{2}$

${v}_{1} = \frac{54 - 9 {v}_{2}}{3}$

Then:

$972 = 3 {\left(\frac{54 - 9 {v}_{2}}{3}\right)}^{2} + 9 {v}_{2}^{2} = {\left(54 - 9 {v}_{2}\right)}^{2} / 3 + 9 {v}_{2}^{2}$

$972 = \frac{2916 - 972 {v}_{2} + 81 {v}_{2}^{2}}{3} + 9 {v}_{2}^{2}$

Let's multiply through by 3 for neatness:

$2916 = 2916 - 972 {v}^{2} + 81 {v}_{2}^{2} + 9 {v}_{2}^{2}$

Rearranging:

$90 {v}_{2}^{2} - 972 {v}_{2} = 0$

Solve using the quadratic formula or otherwise, and you get:

${v}_{2} = \frac{972 \pm \sqrt{944784 - 0}}{180} = \frac{972 \pm 972}{180}$

There are actually two roots, therefore, $0$ and $10.8$ $m {s}^{-} 1$.

That is, the second ball, with a mass of $9$ $k g$ can be either stationary or moving at $10.8$ $m {s}^{-} 1$.

We should solve $\left(1\right)$ with both of these to find the possible values of ${v}_{1}$:

If ${v}_{2} = 0$, ${v}_{1} = \frac{54}{3} = 18 m {s}^{-} 1$

But wait: this is just the condition before the collision! The $3$ $k g$ ball has a velocity of $18$ $m {s}^{-} 1$ and the $9$ $k g$ ball is stationary!

If ${v}_{2} = 10.8$, ${v}_{1} = - 14.4$ $m {s}^{-} 1$.

The minus sign indicates that this velocity is in the opposite direction to the original velocity.

The $3$ $k g$ ball approaches at $18$ $m {s}^{-} 1$ then collides and bounces backward at $14.4$ $m {s}^{-} 1$, propelling the $9$ $k g$ ball forward at $10.8$ $m {s}^{-} 1$.