Jul 15, 2017

The answers are $\left(b\right) = 4.95 m {s}^{-} 1$ and $\left(c\right) = 495 m {s}^{-} 1$

#### Explanation:

Part $\left(a\right)$

When the shot is released, there is a recoil of the gun, but momentum is conserved

${m}_{g} {v}_{g} = - {m}_{b} {v}_{b}$

The gun recoils in the opposie direction.

Part $\left(b\right)$

The height $h = 1.25 m$

Mass of the bullet is $m = 0.005 k g$

Mass of the target is $M = 0.495 k g$

Let the speed of the bullet $+$ target be $= v m {s}^{-} 1$

Then,

$K E = P E$

$\frac{1}{2} \cdot \left(m + M\right) {v}^{2} = \left(m + M\right) g h$

$v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 1.25}$

$= \sqrt{24.5}$

$= 4.95 m {s}^{-} 1$

Part $\left(c\right)$

Let the velocity of the bullet before hitting the target is $= u m {s}^{-} 1$

Then , by the law of conservation of momentum

$\mu + M \cdot 0 = \left(m + M\right) \cdot v$

$u = \frac{m + M}{m} \cdot v$

$= \frac{0.495 + 0.005}{0.005} \cdot 4.95$

$= \frac{0.5}{0.005} \cdot 4.95$

$= 495 m {s}^{-} 1$