Question

A ball with a mass of `4` `kg` and velocity of `6` `ms^(-1)` collides with a second ball with a mass of `8` `kg` and velocity of `- 2` `ms^(-1)`. If `40%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

It helps to mentally imagine what the situation is like. If we take left-to-right as the positive direction, then the `4` `kg` mass is moving left-to-right at `6` `ms^(-1)` while the `8` `kg` mass is moving right-to-left at `2` `ms^(-1)`.

After the collision, the larger mass may be stationary, or moving left-to-right (if it bounces back) with a positive velocity, or moving right-to-left (if it keeps moving in the same direction but more slowly) with a negative velocity.

In this case, momentum is conserved. Kinetic energy is not conserved, but we know how much is lost.

Before the collision:

Momentum: `p = m_1v_1 + m_2v_2 = 4*6 + 8*(-2) = 24 - 16 = 8` `kgms^(-1)` to the right

Kinetic energy:
`E_k = 1/2m_1v_1^2 + 1/2m_2v_2^2=1/2*4*6^2 + 1/2*8*(-2)^2`
`= 72 + 16 = 88` `J`

After the collision:

Momentum will be the same, but we know that 40% of the `E_k` is lost, so 60% remains, so `60%` of `88` = `52.8`J#.

Just for simplicity, I will continue to use `v_1` and `v_2` for the velocities, but these could perhaps be better written as something like `v_(1"after")` and `v_(2"after")`.

`p = 8 = 4v_1 + 8v_2` --> `2 = v_1 + 2v_1`

`E_k = 52.8 = 1/2*4*v_1 + 1/2*8*v_2 = 2v_1 + 4v_2`

We can treat these as a set of simultaneous equations: two equations in two unknowns. Rearranging the first, we can express `v_1` in terms of `v_2`:

`v_1 = 2 - 2v_2`

We can substitute this into the other equation:

`52.8 = 2(2 - 2v_2) + 4v_2 = 4 - 4v_2 + 4v_2 = 4`

Clearly I have done something wrong at some point, since this is mathematical nonsense. I am requesting that other explainers please check and fix this answer.