91Ó°ÊÓ

The total momentum of an isolated system remains the same during a process. The momentum of the constituents of the system may change during the process but the total momentum of system should be unchanged.

According to the law of conservation of momentum as follows:

$m_b{v}_b+m_h{v}_h=m_b{v}_b^{\text{'}}+m_h{v}_h^{\text{'}}......\left(1\right)$

Where,$m_b$is the mass of the bullet,$m_h$is the mass of the hunter,$v_b$and $v_h$are velocities of the bullet and the hunter before firing the rifle,$v_b^{\text{'}}$and$v_h^{\text{'}}$are velocities of the bullet after firing the rifle.

According to the given condition, equation (1) becomes-

$0=m_b{v}_b^{\text{'}}+m_h{v}_h^{\text{'}}......\left(2\right)$

The diagram according to the question, is shown below. Before firing the bullet both the bullet and gun were at rest. After the bullet is fired, both the bullet and the gun move in opposite directions with the velocities$v_h^{\text{'}}$and$v_b^{\text{'}}$respectively.

Before firing the bullet both the bullet and gun were at rest. After the bullet is fired, both the bullet and the gun move in opposite directions with the velocities$v_h^{\text{'}}$and$v_b^{\text{'}}$respectively.

Use the law of conservation of energy:

$$\begin{gathered} 0=m_b{v}_b^{\text{'}}+m_h{v}_h^{\text{'}} \\ {v}_h^{\text{'}}=-\frac{m_b{v}_b^{\text{'}}}{m_h} \\ =-\frac{\left(4.20×{10}^{-3}\right)\mathrm{kg}×965\mathrm{m}/\mathrm{s}}{72.5\mathrm{kg}} \\ =-0.056m/s \end{gathered}$$

Hence, the recoil velocity of the hunter is 0.056 m/s, opposite to the direction of motion of bullet.

The diagram according to the question,

Before firing the bullet both the bullet and gun were at rest. After the bullet is fired making an angle with the horizontal, a component of the recoil of gun will push the hunter backward.

Using the law of conservative of energy, we have-

$$\begin{gathered} 0=m_b{v}_b^{\text{'}}\cos \mathrm{Î\pm }+m_h{v}_h^{\text{'}} \\ {m}_h{v}_h^{\text{'}}=-m_b{v}_b^{\text{'}}\cos \mathrm{Î\pm } \\ {v}_b^{\text{'}}=-\frac{m_b{v}_b^{\text{'}}}{m_h}\cos \mathrm{Î\pm } \end{gathered}$$

Substituting the given values in the above equation,

$$\begin{gathered} =-\frac{\left(4.20×{10}^{-3}\right)\mathrm{kg}×965\mathrm{m}/\mathrm{s}}{72.5\mathrm{kg}}×0.5592 \\ =-0.031\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the recoil velocity of the hunter, if he fires the rifle at from the horizontal, is 0.031 m/s.