Question

A ball with a mass of `1 kg` and velocity of `7 m/s` collides with a second ball with a mass of `6 kg` and velocity of `- 4 m/s`. If `40%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

`v_1 = -4.62 m/s and v_2 = -2.06 m/s`

Explanation:

KE initial= `1/2m_1*u_1^2+1/2m_2*u_2^2`
KE initial= `1/2(1kg)(7m/s)^2+1/2(6kg)(-4m/s)^2`
KE initial= 72.5 J

If 60% KE is conserved then 72.5*.6= 43.5 J is KE final

Another consideration: momentum, p, must be conserved.

Momentum before = momentum after

`p = m_1*u_1 + m_2*u_2 = m_1*v_1 + m_2*v_2`

`(1kg)(7m/s)+(6kg)(-4m/s) = 1 kg*v_1 + 6 kg*v_2`

`7 (kg*m)/s - 24 (kg*m)/s = 1 kg*v_1 + 6 kg*v_2`

`-17 (kg*m)/s = 1 kg*v_1 + 6 kg*v_2`

Solving for `v_1`,

`v_1 = (-17 (kg*m)/s - 6 kg*v_2)/(1 kg) = -17 m/s - 6*v_2`

Going back to the kinetic energy

`1/2m_1v_1^2+1/2m_2v_2^2=` KE final which, from above, is 43.5 J.

`1/2(1kg)v_1^2+1/2(6kg)v_2^2= 43.5 J`

Plugging in the expression for `v_1` from the momentum analysis,

`1/2(1kg)(-17 m/s - 6*v_2)^2+1/2(6kg)v_2^2= 43.5 J`

`1/2(1kg)(17^2 m^2/s^2 + 2*17 m/s*6*v_2 + 6^2*v_2^2)+1/2(6kg)v_2^2= 43.5 J`

`1/2(1kg)(289 m^2/s^2 + 204 m/s*v_2 + 36*v_2^2)+1/2(6kg)v_2^2= 43.5 J`

Now I will multiply thru by 2, multiply thru the first set of parentheses by that 1 kg, and substitute `J` for the combination of units `(kg*m^2)/s^2`,

`289 J + 204 (kg*m)/s*v_2 + 36 kg*v_2^2+6kg*v_2^2= 87 J`

Now, notice the units in that equation. It is clear that all terms are energy and that `v_2` will be in units of m/s when the quadratic equation is solved. Therefore I will drop the units for simplicity.

`289 + 204 *v_2 + 36 *v_2^2+6kg*v_2^2= 87`

`42 *v_2^2 + 204 *v_2+ 202 = 0`

My working of the quadratic equation yields 2 values and neither is obviously invalid.

`v_1 = -4.62 m/s, -0.24 m/s`

Let's see what value for `v_2` each gives. Using the last equation from the momentum study again,

`-4.62 m/s = -17 m/s - 6*v_2`

`- 6*v_2 = -4.62 m/s + 17 m/s`

`v_2 = (-4.62 m/s + 17 m/s)/-6 = -2.06 m/s`

Repeating with the other result from the quadratic equation work,

`-0.24 m/s = -17 m/s - 6*v_2`

`- 6*v_2 = -0.24 m/s + 17 m/s`

`v_2 = (-0.24 m/s + 17 m/s)/-6 = -2.49 m/s`

I expected a clear-cut way to eliminate one of those results. I do notice that both balls are going the direction that the second ball was going before the collision. Therefore the solution that has `v_1 < v_2` (smaller one running away from the big one) looks better.

I will post this now but review my work and ask for a double check.

I hope this helps,
Steve