Q106 / 180 · AIIMS 2018 (26 May) · Physics

A bomb moving with velocity $(40 \hat{i}+50 \hat{j}-25 \hat{k}) \mathrm{m} \mathrm{s}^{-1}$ explodes into two pieces of mass ratio $1: 4$. After explosion the smaller piece moves away with velocity $(200 \hat{i}+70 \hat{j}+15 \hat{k}) \mathrm{m} \mathrm{s}^{-1}$. The velocity of larger piece after explosion is

  • Correct answer
    A
    $45 \hat{j}-35 \hat{k}$
  • B
    $45 \hat{i}-35 \hat{j}$
  • C
    $45 \hat{k}-35 \hat{j}$
  • D
    $-35 \hat{i}+45 \hat{k}$

Solution

( Let the mass of the unexploded bomb be $5 \mathrm{~m}$. It explodes into the two pieces of masses $m$ and $4 m$ respectively. Initial momentum of the unexploded bomb $=5 m(40 \hat{i}+50 \hat{j}-25 \hat{k})$ After explosion, momentum of the smaller piece $=m \vec{v}_1=m(200 \hat{i}+70 \hat{j}+15 \hat{k})$ and momentum of the larger piece $=4 m \vec{v}_2$ where $\vec{v}_1$ and $\vec{v}_2$ are the velocities of the two pieces respectively. According to the law of conservation of momentum, we get $\begin{aligned} & 5 m(40 \hat{i}+50 \hat{j}-25 \hat{k})=m(200 \hat{i}+70 \hat{j}+15 \hat{k})+4 m \vec{v}_2 \\ & 4 m \vec{v}_2=5 m(40 \hat{i}+50 \hat{j}-25 \hat{k})-m(200 \hat{i}+70 \hat{j}+15 \hat{k}) \\ & \quad \vec{v}_2=\frac{1}{4}(180 \hat{j}-140 \hat{k})=45 \hat{j}-35 \hat{k}\end{aligned}$

Hence, the correct answer is option: 'A'.

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