Question

1. A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20-g bullets at 965m/s. The mass of the hunter (including his gun) is 72.5kg, and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0° above the horizontal.

Answer 1

Answer on Question #84444 - Physics - Mechanics | Relativity

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20-g bullets at $965\mathrm{m / s}$ . The mass of the hunter (including his gun) is $72.5\mathrm{kg}$ , and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at $56.0{}^{\circ}$ above the horizontal.

Solution.

According to the law of momentum conservation:

$$m _ {b} \vec {\vartheta} _ {b} + m _ {h} \vec {\vartheta} _ {h} = m _ {b} \vec {\vartheta} _ {b} ^ {\prime} + m _ {h} \vec {\vartheta} _ {h} ^ {\prime},$$

where $m_{b}$ is the mass of the bullet, $m_{h}$ is the mass of the hunter, $\vec{\vartheta}_{b}$ and $\vec{\vartheta}_{h}$ are velocities of the bullet and the hunter before firing the rifle, $\vec{\vartheta}_{b}^{\prime}$ and $\vec{\vartheta}_{h}^{\prime}$ are velocities of the bullet and the hunter after firing the rifle.

Taking into account, that $\vec{\vartheta}_h' = \vec{\vartheta}_b' = 0$ , we receive:

$$0 = m _ {b} \vec {\vartheta} _ {b} ^ {\prime} + m _ {h} \vec {\vartheta} _ {h} ^ {\prime}.$$

Let's rewrite (2) in scalar form.

Case (a):

$$0 = m _ {b} \vartheta_ {b} ^ {\prime} - m _ {h} \vartheta_ {h} ^ {\prime}$$ $$m _ {h} \vartheta_ {h} ^ {\prime} = m _ {b} \vartheta_ {b} ^ {\prime}$$ $$\vartheta_ {h} ^ {\prime} = \frac {m _ {b} \vartheta_ {b} ^ {\prime}}{m _ {h}}$$ $$\begin{array}{l} \vartheta_ {h} ^ {\prime} = \frac {4 . 2 0 \cdot 1 0 ^ {- 3} \cdot 9 6 5}{7 2 . 5} = 0. 0 5 5 9 (m / s) = \\ = 5 5. 9 (m m / s) \\ \end{array}$$

Case (b):

$$0 = m _ {b} \vartheta_ {b} ^ {\prime} \cos \alpha - m _ {h} \vartheta_ {h} ^ {\prime}$$ $$m _ {h} \vartheta_ {h} ^ {\prime} = m _ {b} \vartheta_ {b} ^ {\prime} \cos \alpha$$ $$\vartheta_ {h} ^ {\prime} = \frac {m _ {b} \vartheta_ {b} ^ {\prime}}{m _ {h}} \cos \alpha$$ $$\begin{array}{l} \vartheta_ {h} ^ {\prime} = \frac {4 . 2 0 \cdot 1 0 ^ {- 3} \cdot 9 6 5}{7 2 . 5} \cdot 0. 5 5 9 2 = \\ = 0. 0 3 1 (m / s) = 3 1 (m m / s) \\ \end{array}$$

Answer: (a) $\vartheta_h' = 55.9 \, \text{mm/s}$ ; (b) $\vartheta_h' = 31 \, \text{mm/s}$ .

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