Question

# A bullet of mass $20g$ is fired horizontally with a velocity of $150m/s$ from a pistol of mass $2Kg$. Find the recoil velocity of the pistol.

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Solution

## Step 1: GivenMass of the bullet, ${m}_{1}=20g=0.02Kg$Mass of the pistol, ${m}_{2}=2Kg$The initial velocity of the bullet and the pistol, ${u}_{1}={u}_{2}=0$The final velocity of the bullet ${v}_{1}=150m/s$Let the recoil velocity of the pistol be ${v}_{2}$Step 2: Find the total momentumsAccording to the Law of conservation of momentum, the initial momentum will be equal to the final momentum.The total momentum before the bullet is fired (${M}_{i}$) will be zero as both the bullet and the pistol are at rest.The total momentum after the bullet is fired (${M}_{f}$) will be,${M}_{f}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}\phantom{\rule{0ex}{0ex}}=0.02\left(150\right)+2\left({v}_{2}\right)\phantom{\rule{0ex}{0ex}}=3+2{v}_{2}$Step 3: Equate the momentumsThe Law of conservation of momentum states that the initial momentum will be equal to the final momentum. $⇒{M}_{f}={M}_{i}$$3+2{v}_{2}=0\phantom{\rule{0ex}{0ex}}{v}_{2}=\frac{-3}{2}\phantom{\rule{0ex}{0ex}}{v}_{2}=-1.5m/s$The value is negative as it is a recoil velocity. The direction of recoil velocity is opposite to the direction of motion.Hence, the recoil velocity of the pistol is $1.5m/s$.

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