Answer to Question #158105 in Mechanics | Relativity for aleyna

Question

A 7-kg rifle is used to fire a 10-g bullet that travels with a speed of 700 m/s. (a) What is the speed of recoil of the rifle? (b) How much energy does it transmit to the shoulder of the person using the rifle as it stops?

Result: (a) -1.0 m/s (b) 35 J

Answer 1

a) Let's use the momentum conservation law:

$$0 = m_{b}v_b +m_rv_r$$

where stands for the zero velocities of the rifle and bullet before the fire, $m_b = 10g = 0.01 kg, m_r = 7kg$ are the masses of the bullet and rifle respectively, $v_b = 700m/s, v_r$ are the velocities of the bullet and rifle after the fire respoctively. Thus, obtain:

$$v_r =- \dfrac{m_bv_b}{m_r}\\ v_r = -\dfrac{0.01\cdot 700}{7} = -1m/s$$

b) The rifle should transmit all its kinetic energy to stop. The kinetic energy is given as follows:

$$E_k = \dfrac{m_rv_r^2}{2}\\ E_k = \dfrac{7\cdot 1^2}{2} = 3.5J$$

Answer. a) -1m/s, b) 3.5J.