An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 \mathrm{~kg}$ moves with a speed of $12 \mathrm{~ms}^{-1}$ and the second part of mass $2 \mathrm{~kg}$ moves with $8 \mathrm{~ms}^{-1}$ speed. If the…

Question

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 \mathrm{~kg}$ moves with a speed of $12 \mathrm{~ms}^{-1}$ and the second part of mass $2 \mathrm{~kg}$ moves with $8 \mathrm{~ms}^{-1}$ speed. If the third part flies off with $4 \mathrm{~ms}^{-1}$ speed, then its mass is

Aicharya · Accepted Answer

We have $p_1+p_2+p_3=0 [\because p=m v]$
$\begin{aligned}
& \therefore 1 \times 12 i+2 \times 8 j+p_3=0 \\
& \Rightarrow 12 \mathrm{i}+16 \mathrm{j}+\mathrm{p}_3=0 \\
& \Rightarrow \mathrm{P}_3=-(12 \mathrm{i}+16 \mathrm{j}) \\
& \therefore \mathrm{p}_3=\sqrt{(12)^2+(16)^2} \\
& =\sqrt{144+256} \\
& =20 \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}$
Now, $p_3=m_3 v_3$
$\Rightarrow m_3=\frac{p_3}{V_3}=\frac{20}{4}=5 \mathrm{~kg}$

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to…

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