First slide

Third law

Question

A shell of mass 200 g is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ${ms}^{- 1}$ , then the recoil speed of the gun is

Moderate

A

16 $cm s^{- 1}$
B

8 $cm s^{- 1}$
C

8 ${ms}^{- 1}$
D

16 ${ms}^{- 1}$

Solution

Here,
Mass of the shell, m = 200g = 200 $\times$ $10^{- 3} kg$
Mass of the gun, M = 100 kg
Muzzle speed of the shell, V = 80 ${ms}^{- 1}$
Recoil speed of the gun, v = ?
According to the principle of conservation of linear momentum,
mV + Mv = 0
$v = \frac{mV}{M} = - \frac{(200 \times 10^{- 3} kg) (80 {ms}^{- 1})}{100 kg}$
= -16 $\times 10^{- 2} {ms}^{- 1} = - 16 cm s^{- 1}$
-ve sign shows that the recoil speed of the gun will be in a direction opposite to that of the shell.

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