Question

A 30-06 caliber hunting rifle fires a bullet of mass 0.0149 kg with a velocity of 430 m/s to the right. The rifle has a mass of 4.9 kg.

a) What is the recoil speed of the rifle as the bullet leaves the rifle?

Answer in units of m/s.

b) If the rifle is stopped by the hunter’s shoulder in a distance of 1.38 cm, what is the mag- nitude of the average force exerted on the shoulder by the rifle?

Answer in units of N.

Answer 1

Applying conservation of momentum, (there is no external force)

Initial momentum = final momentum

initially both bullet(b) and Rifle(r) are in rest before firing the bullet, finally bullet goes too right and rifle will recoil to left

U_b=0 m/s U_r = 0 m/s   V_b= 430 m/s V_r = ?

m_bU_b + m_rU_r = m_bV_b - m_rV_r   (negative sign indicates the direction is opposite for recoil)

0 + 0 = (0.0149 * 430) - (4.9*V_r)

(4.9*V_r) = (0.0149 * 430)

V_r= (0.0149 * 430)/4.9

V_r = 1.307 m/s (recoil speed of the rifle)

b) distance of stop, d = 1.38 cm = 0.0138 m

Force exerted on the shoulder, F = m_r * a

V²_r - U²_r =2 * a* d

a = (V²_r - 0)/ (2*d) = 1.709/0.0276 = 61.81

F= 4.9 * 61.81 =302.87 N