# A ball with a mass of  3 kg is rolling at 4 ms^-1 and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Jan 15, 2016

In an elastic collision, both momentum and kinetic energy are conserved. The velocity of the $3 k g$ ball is $0.9 \mathmr{and} 2.7 m {s}^{-} 1$ and the velocity of the $4 k g$ ball is $2.7 \mathmr{and} 1.1 m {s}^{-} 1$.

#### Explanation:

Momentum is conserved in all collisions. Kinetic energy is conserved in elastic collisions but not in inelastic or partially elastic collisions.

The initial momentum of the whole system is $p = m v$ for the $3 k g$ ball: the $4 k g$ ball has zero momentum because its has zero velocity.

$p = m v = 3 \cdot 4 = 12 k g m {s}^{-} 1$

The total momentum after the collision will be the same after the collision. If we call the $3 k g$ ball 1 and the $4 k g$ ball 2, the final momentum will be given by:

$p = {m}_{1} {v}_{1} + {m}_{2} {v}_{2} = 3 {v}_{1} + 4 {v}_{2} = 12$ (call this Equation 1)

The total kinetic energy before the collision will be ${E}_{k} = \frac{1}{2} m {v}^{2}$ for the $3 k g$ ball only - the $4 k g$ ball has zero kinetic energy because it has zero velocity.

${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 3 \cdot {4}^{2} = 24 J$

Since kinetic energy is conserved, the final kinetic energy will be the same, and will be given by:

${E}_{k} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{3}{2} {v}_{1}^{2} + \frac{4}{2} {v}_{2}^{2} = 24$

Multiply both sides by 2 to make it tidier:

$3 {v}_{1}^{2} + 4 {v}_{2}^{2} = 48$ (call this Equation 2)

We now have two equations and two unknowns, so we can solve them as simultaneous equations. Rearranging Equation 1 to express ${v}_{2}$ in terms of ${v}_{1}$:

${v}_{2} = \frac{12 - 3 {v}_{1}}{4}$

Substituting this into Equation 2:

$3 {v}_{1}^{2} + 4 {\left(\frac{12 - 3 {v}_{1}}{4}\right)}^{2} = 48$

I'll leave the algebra as an exercise for the reader, but this solves to give ${v}_{1} = 0.9 m {s}^{-} 1 \mathmr{and} 2.5 m {s}^{-} 1$. Substituting this back into Equation 1 gives ${v}_{2} = 2.7 m {s}^{-} 1 \mathmr{and} 1.1 m {s}^{-} 1$, respectively.

Substituting these values should confirm that both momentum and kinetic energy were conserved.