# 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?

Feb 21, 2018

${v}_{1} = - 4.62 \frac{m}{s} \mathmr{and} {v}_{2} = - 2.06 \frac{m}{s}$

#### Explanation:

KE initial= $\frac{1}{2} {m}_{1} \cdot {u}_{1}^{2} + \frac{1}{2} {m}_{2} \cdot {u}_{2}^{2}$
KE initial= $\frac{1}{2} \left(1 k g\right) {\left(7 \frac{m}{s}\right)}^{2} + \frac{1}{2} \left(6 k g\right) {\left(- 4 \frac{m}{s}\right)}^{2}$
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

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

$\left(1 k g\right) \left(7 \frac{m}{s}\right) + \left(6 k g\right) \left(- 4 \frac{m}{s}\right) = 1 k g \cdot {v}_{1} + 6 k g \cdot {v}_{2}$

$7 \frac{k g \cdot m}{s} - 24 \frac{k g \cdot m}{s} = 1 k g \cdot {v}_{1} + 6 k g \cdot {v}_{2}$

$- 17 \frac{k g \cdot m}{s} = 1 k g \cdot {v}_{1} + 6 k g \cdot {v}_{2}$

Solving for ${v}_{1}$,

${v}_{1} = \frac{- 17 \frac{k g \cdot m}{s} - 6 k g \cdot {v}_{2}}{1 k g} = - 17 \frac{m}{s} - 6 \cdot {v}_{2}$

Going back to the kinetic energy

$\frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} =$ KE final which, from above, is 43.5 J.

$\frac{1}{2} \left(1 k g\right) {v}_{1}^{2} + \frac{1}{2} \left(6 k g\right) {v}_{2}^{2} = 43.5 J$

Plugging in the expression for ${v}_{1}$ from the momentum analysis,

$\frac{1}{2} \left(1 k g\right) {\left(- 17 \frac{m}{s} - 6 \cdot {v}_{2}\right)}^{2} + \frac{1}{2} \left(6 k g\right) {v}_{2}^{2} = 43.5 J$

$\frac{1}{2} \left(1 k g\right) \left({17}^{2} {m}^{2} / {s}^{2} + 2 \cdot 17 \frac{m}{s} \cdot 6 \cdot {v}_{2} + {6}^{2} \cdot {v}_{2}^{2}\right) + \frac{1}{2} \left(6 k g\right) {v}_{2}^{2} = 43.5 J$

$\frac{1}{2} \left(1 k g\right) \left(289 {m}^{2} / {s}^{2} + 204 \frac{m}{s} \cdot {v}_{2} + 36 \cdot {v}_{2}^{2}\right) + \frac{1}{2} \left(6 k g\right) {v}_{2}^{2} = 43.5 J$

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 $\frac{k g \cdot {m}^{2}}{s} ^ 2$,

$289 J + 204 \frac{k g \cdot m}{s} \cdot {v}_{2} + 36 k g \cdot {v}_{2}^{2} + 6 k g \cdot {v}_{2}^{2} = 87 J$

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.

$289 + 204 \cdot {v}_{2} + 36 \cdot {v}_{2}^{2} + 6 k g \cdot {v}_{2}^{2} = 87$

$42 \cdot {v}_{2}^{2} + 204 \cdot {v}_{2} + 202 = 0$

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

${v}_{1} = - 4.62 \frac{m}{s} , - 0.24 \frac{m}{s}$

Let's see what value for ${v}_{2}$ each gives. Using the last equation from the momentum study again,

$- 4.62 \frac{m}{s} = - 17 \frac{m}{s} - 6 \cdot {v}_{2}$

$- 6 \cdot {v}_{2} = - 4.62 \frac{m}{s} + 17 \frac{m}{s}$

${v}_{2} = \frac{- 4.62 \frac{m}{s} + 17 \frac{m}{s}}{-} 6 = - 2.06 \frac{m}{s}$

Repeating with the other result from the quadratic equation work,

$- 0.24 \frac{m}{s} = - 17 \frac{m}{s} - 6 \cdot {v}_{2}$

$- 6 \cdot {v}_{2} = - 0.24 \frac{m}{s} + 17 \frac{m}{s}$

${v}_{2} = \frac{- 0.24 \frac{m}{s} + 17 \frac{m}{s}}{-} 6 = - 2.49 \frac{m}{s}$

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