Question

A ball with a mass of `6` `kg` and velocity of `4` `ms^-1` collides with a second ball with a mass of `4` `kg` and velocity of `- 5` `ms^-1`. If `75%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

In this instance momentum is conserved. Kinetic energy is not but we know how much was lost.

`v_1 = 1.9` `ms^-1` or `1.3` `ms^-1`

`v_2 = -1.85` `ms^-1` or `-0.95` `ms^-1`

Explanation:

Before the collision

Momentum:

`p = m_1v_1 + m_2v_2 = 6 \times 4 + 4 \times-5 = 4` `kgms^-1`

Kinetic energy:

`E_k = 1/2 m_1v_1^2 + 1/2m_2v_2^2 = 1/2 \times 6 \times 4^2 + 1/2 \times 4 \times (-5)^2 = 98` `J`

After the collision

(note that the velocities now are different, even though I have still called the `v_1` and `v_2`. I've done this for simplicity in writing it out, but you need to keep track of which velocities we are talking about at any given time. From here on in, these are the final velocities: up to now we have been talking about the initial velocities.)

Momentum:

`p = m_1v_1 + m_2v_2 = 6 v_1 + 4 v_2`

Kinetic energy:

`E_k = 1/2 m_1v_1^2 + 1/2m_2v_2^2 = 1/2 \times 6 \times v_1^2 + 1/2 \times 4 \times v_2^2 = 3v_1^2 + 2v_2^2`

Now, we know that momentum is conserved, so `p` before equals `p` after, and we can make an equation:

`6 v_1 + 4 v_2 = 4` : call this Equation 1

We also know that `75%` of the initial kinetic energy was lost, so we only have `25%` left. `25%` of `98` `J` is `24.5` `J`, and we can equate this to the kinetic energy after the collision:

`3v_1^2 + 2v_2^2 = 24.5` : call this Equation 2.

Now we have two equations in two unknowns, so we can use our tools for 'simultaneous equations' to solve for `v_1` and `v_2`. There are a number of possible approaches, this is just one of them:

Let's divide Equation 1 by 4:

`3/2v_1 + v_2 = 1`

Rearranging to make `v_2` the subject:

`v_2 = 1 - 3/2v_1`

We can substitute this expression for `v_2` into Equation 2, and that will give us an equation that is only in terms of `v_1`, which we can solve:

`3v_1^2 + 2(1 - 3/2v_1)^2 = 24.5`

Expanding the brackets:

`3v_1^2 + 2(1 - 3v_1 + 9/4v_1^2) = 24.5`

`3v_1^2 + 2 - 6v_1 + 9/2v_1^2 = 24.5`

`9/2v_1^2 - 3v_1 + 2 = 24.5`

This is a quadratic equation, which we can solve using the quadratic formula or other methods, but first we need to make the right side equal zero, so we subtract 24.5 from both sides:

`9/2v_1^2 - 3v_1 -22.5 = 0`

Solving it yields:

`v_1 = 1.9` `ms^-1` or `1.3` `ms^-1`

We can substitute these values back in to our rearranged Equation 1 to find the values of `v_2`:

`v_2 = 1 - 3/2v_1`

So `v_2 = -1.85` `ms^-1` or `-0.95` `ms^-1`

The quadratic equation yields two solutions, corresponding to two possible outcomes of the collision.