# A ball with a mass of 5 kg moving at 9 m/s hits a still ball with a mass of 8 kg. If the first ball stops moving, how fast is the second ball moving?

Mar 17, 2018

The velocity of the second ball after the collision is $= 5.625 m {s}^{-} 1$

#### Explanation:

We have conservation of momentum

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

The mass the first ball is ${m}_{1} = 5 k g$

The velocity of the first ball before the collision is ${u}_{1} = 9 m {s}^{-} 1$

The mass of the second ball is ${m}_{2} = 8 k g$

The velocity of the second ball before the collision is ${u}_{2} = 0 m {s}^{-} 1$

The velocity of the first ball after the collision is ${v}_{1} = 0 m {s}^{-} 1$

Therefore,

$5 \cdot 9 + 8 \cdot 0 = 5 \cdot 0 + 8 \cdot {v}_{2}$

$8 {v}_{2} = 45$

${v}_{2} = \frac{45}{8} = 5.625 m {s}^{-} 1$

The velocity of the second ball after the collision is ${v}_{2} = 5.625 m {s}^{-} 1$

Mar 17, 2018

Initial momentum of the system was 5×9+8×0 Kgms^-2

After the collision momentum was 5×0+8×v Kgms^-2 where,$v$ is the velocity of the 2nd ball after collision.

So,applying law of conservation of momentum we get,

$45 = 8 v$

Or, $v = 5.625 m {s}^{-} 1$