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

Mar 13, 2016

Intuition and experience tell us all the momentum and energy will transfer from one ball to the other, but we should calculate to check.

The detailed calculation below shows that, indeed, after the collision the velocity of the first ball is $0$ and that of the second ball is $3$ $m {s}^{-} 1$.

Explanation:

An elastic collision means that both momentum, $p = m v$, and kinetic energy, ${E}_{k} = \frac{1}{2} m {v}^{2}$, are conserved.

It might be simpler if we call the $3$ $k g$ ball that was initially moving '1' and the one that was initially stationary '2', just to make it easier to refer to them.

Momentum before the collision:

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

Momentum after the collision:

$p = {m}_{1} {v}_{1} + {m}_{2} {v}_{2} = 3 \cdot {v}_{1} + 3 \cdot {v}_{2}$

Conservation of momentum tells us these two things are equal to each other, so:

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

Divide through by 3 to make it simpler:

${v}_{1} + {v}_{2} = 3$   call this Equation (1)

That's one equation in two unknowns. We'll need another equation if we're going to be able to solve this.

Kinetic energy before the collision:

${E}_{k} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{1}{2} \cdot 3 \cdot {3}^{2} + \frac{1}{2} \cdot 3 \cdot {0}^{2} = \frac{27}{2}$ $J$

Kinetic energy after the collision:

${E}_{k} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{3}{2} {v}_{1}^{2} + \frac{3}{2} {v}_{2}^{2}$ $J$

Kinetic energy is not always conserved, but in this case we are told that collision is elastic, which means that kinetic energy is conserved. That means these two things - before and after - are also equal to each other, so:

$\frac{3}{2} {v}_{1}^{2} + \frac{3}{2} {v}_{2}^{2} = \frac{27}{2}$

Let's divide through by 3/2 to make it neater:

${v}_{1}^{2} + {v}_{2}^{2} = 9$   call this Equation (2)

By rearranging Equation (1) we can get an expression for ${v}_{2}$:

${v}_{2} = 3 - {v}_{1}$

Substitute this into Equation 2 and solve:

${v}_{1}^{2} + {\left(3 - {v}_{1}\right)}^{2} = 9 = {v}_{1}^{2} + 9 - 6 {v}_{1} + {v}_{1}^{2}$

Subtracting 9 from both sides and rearranging:

$2 {v}_{1}^{2} - 6 {v}_{1} = 0$

Solve this quadratic equation using the quadratic formula or another method and you will find the roots:

${v}_{1} = 3 \mathmr{and} {v}_{1} = 0$

Fascinatingly, these two roots relate to the situation before and after the collision! Initially the first ball had a velocity of 3, but after the collision it had a velocity of 0.

Substituting ${v}_{1} = 0$ back into Equation (1), we find that ${v}_{2} = 3$.

(this is the condition after the collision. If we had substituted ${v}_{1} = 3$ we would have found ${v}_{2} = 0$, the situation before the collision)