# A ball with a mass of  2 kg is rolling at 5 m/s and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

Mar 12, 2016

#### Answer:

$\implies u = \frac{5}{3} m {s}^{-} 1$ and $v = \frac{20}{27} m {s}^{-} 1$

#### Explanation:

If post collision velocities be u m/s for 1st one and v m/s for 2nd one ,then applying conservation of momentum
$u \times 2 + v \times 9 = 2 \times 5 + 0 \times 9$
$\implies u \times 2 + v \times 9 = 10$
$\implies v \times 9 = 10 - u \times 2$
$\implies v \times 9 = \left(5 - u\right) \times 2 \ldots \ldots . \left(1\right)$

Now collision being elastic KE will be conserved and consequently the relative velocity of 1st object with respect to 2nd is reversed .
So $\frac{1}{2} \times 2 \times {5}^{2} + \frac{1}{2} \times 9 \times {0}^{2} = \frac{1}{2} \times 2 \times {u}^{2} + \frac{1}{2} \times 9 {v}^{2}$
$\implies 2 \times {5}^{2} = 2 \times {u}^{2} + 9 {v}^{2}$
$\implies 2 \times {5}^{2} - 2 \times {u}^{2} = 9 {v}^{2}$
$\implies 2 \times \left({5}^{2} - {u}^{2}\right) = 9 {v}^{2}$
$\implies 2 \times \left(5 + u\right) \left(5 - u\right) = 9 {v}^{2}$
$\implies \left(5 + u\right) 9 v = 9 {v}^{2}$
$\implies \left(5 + u\right) = 9 v \ldots \ldots \left(2\right)$ $\left[\therefore v \ne 0\right]$
Comparing (1) and (2)
$5 + u = 10 - 2 u$
$\implies 3 u = 10 - 5$
$\implies u = \frac{5}{3} m {s}^{-} 1$
and $v = \frac{20}{27} m {s}^{-} 1$