A bullet of mass 20 g is fired horizontally with a velocity of 100 m/s from a pistol of mass 1.5 kg. Calculate the recoil velocity of the pistol.

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Solution

Step 1: Given data
Mass of bullet (M_B) = 20 g
We know that $1 k g = 1000 g$
Mass of pistol (M_P) = 1.5 kg
= $1.5 \times 10^{3}$ g
Velocity of bullet (V_B) = 100 m/s
Step 2: Conservation of linear momentum
According to momentum conservation law, the total initial momentum $(T_{i})$ must be equal to the final momentum $(T_{f}) .$ That is;
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Step 3: Finding the recoil velocity
Initially both pistol and bullet are at rest. So, the initial velocities of the object will be zero.
Let the recoil velocity of the pistol be ‘V’. By conserving the momentum, we get;
Total momentum after the fire = Total momentum before the fire
$M_{B} U_{B} + M_{P} U_{P} = M_{B} V_{B} + M_{P} V 0 = M_{B} V_{B} + M_{P} V M_{B} V_{B} = - M_{P} V 20 g \times 100 m s^{- 1} = - (1.5 \times 10^{3} g) V V = - \frac{20 \times 100}{1.5 \times 1000} V = - 1.33 m s^{- 1}$
Hence, the recoil velocity of pistol is $- 1.33 m s^{- 1}$ and the negative sign shows that both velocities are in opposite directions.

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