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 hunterâ€™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_{b}U_{b} + m_{r}U_{r} =
m_{b}V_{b} - m_{r}V_{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

V^{2}_{r} - U^{2}_{r} =2 * a*
d

a = (V^{2}_{r} - 0)/ (2*d) = 1.709/0.0276 =
61.81

F= 4.9 * 61.81 =302.87 N

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