Question

# A shell of mass $m$ moving with velocity $V$ suddenly breaks into 2 pieces. The part having mass $\frac{m}{4}$ remains stationary. The velocity of the other shell will be:

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Solution

## Step 1: Given dataThe mass of the body is $m$The mass of the two broken parts after the explosion is $\frac{m}{4}$ and $\left(m-\frac{m}{4}\right)$.The velocity of the body is $V$.The velocity of the first piece of mass $\frac{m}{4}$ is ${v}_{1}=0$.Step 2: Formula usedThe linear momentum of a body is the product of the mass and velocity of the body, i.e, linear momentum, $P=mv$, where, m and v are the mass and velocity of a body.We know in an explosion the linear momentum is conserved. So, $MV={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$, where, M and V are the mass of the body, ${m}_{1},{m}_{2}$ and ${v}_{1},{v}_{2}$ are the masses and velocities of the broken pieces.Step 3: Finding the velocity of the second pieceLet, ${v}_{2}$ be the velocity of the second piece.According to the conservation of linear momentum, $MV={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$So, $mv=\left(\frac{m}{4}×0\right)+\left(m-\frac{m}{4}\right){v}_{2}\left(Since,{v}_{1}=0\right).\phantom{\rule{0ex}{0ex}}ormv=\frac{3m}{4}×{v}_{2}\phantom{\rule{0ex}{0ex}}or{v}_{2}=\frac{4mv}{3m}=\frac{4}{3}v\phantom{\rule{0ex}{0ex}}or{v}_{2}=\frac{4}{3}v.$Therefore, the velocity of the second piece is $\frac{4}{3}v$.

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