# A ball with a mass of 1 kg  and velocity of 2 m/s collides with a second ball with a mass of 3 kg and velocity of - 4 m/s. If 20% of the kinetic energy is lost, what are the final velocities of the balls?

Aug 13, 2018

The solution is $= \left(\left(- 1.03 , - 3.68\right)\right) m {s}^{-} 1$

#### Explanation:

There is conservation of momentum

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} x + {m}_{2} y$

Plugging in the above values

$1 \cdot 2 + 3 \cdot - 4 = 1 x + 3 y$

$x + 3 y = 2 - 12 = - 10$

$x + 3 y = - 10$.................................$\left(1\right)$

20% of kinetic energy is lost

$\left(\frac{1}{2} {m}_{1} {u}_{1}^{2} + \frac{1}{2} {m}_{2} {u}_{2}^{2}\right) \cdot \frac{8}{10} = \frac{1}{2} {m}_{1} {x}^{2} + \frac{1}{2} {m}_{2} {y}^{2}$

Plugging the data

$\left(1 \cdot {2}^{2} + 3 \cdot {\left(- 4\right)}^{2}\right) \cdot \frac{8}{10} = 1 {x}^{2} + 3 {y}^{2}$

${x}^{2} + 3 {y}^{2} = 41.6$

${x}^{2} + 3 {y}^{2} = 41.6$......................................$\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ graphically

graph{(x+3y+10)(x^2+3y^2-41.6)=0 [-18.02, 18.03, -9.01, 9.01]}

The solutions are $= \left(- 1.03 , - 3.68\right)$ or $= \left(- 6.03 , - 1.32\right)$