# A ball with a mass of 9 kg  and velocity of 2 m/s collides with a second ball with a mass of 3 kg and velocity of - 5 m/s. If 10% of the kinetic energy is lost, what are the final velocities of the balls?

Jul 29, 2018

The solution is $= \left(\left(- 1.41 , 5.23\right)\right) m {s}^{-} 1$

#### 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

$9 \cdot 2 + 3 \cdot - 5 = 9 x + 3 y$

$9 x + 3 y = 18 - 15 = 3$

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

10% of kinetic energy is lost

$\left(\frac{1}{2} {m}_{1} {u}_{1}^{2} + \frac{1}{2} {m}_{2} {u}_{2}^{2}\right) \cdot \frac{9}{10} = \frac{1}{2} {m}_{1} {x}^{2} + \frac{1}{2} {m}_{2} {y}^{2}$

Plugging the data

$\left(9 \cdot {2}^{2} + 3 \cdot {5}^{2}\right) \cdot \frac{9}{10} = 9 {x}^{2} + 3 {y}^{2}$

$9 {x}^{2} + 3 {y}^{2} = 99.9$

$3 {x}^{2} + {y}^{2} = 33.3$......................................$\left(2\right)$

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

graph{(3x+y-1)(3x^2+y^2-33.3)=0 [-18.02, 18.03, -9.01, 9.01]}

The solutions are $= \left(- 1.41 , 5.23\right)$ or $= \left(1.91 , - 4.73\right)$