An explosion blows a stationary rock into three parts. Two parts of masses \(1 \mathrm{~kg}\) and \(2…
An explosion blows a stationary rock into three parts. Two parts of masses \(1 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) moves at right angles to one another with velocities \(12 \mathrm{~ms}^{-1}\) and \(8 \mathrm{~ms}^{-1}\), respectively. If the velocity of third part is 4 \(\mathrm{ms}^{-1}\), the mass of the rock is
\(8 \mathrm{~kg}\)
\(5 \mathrm{~kg}\)
\(17 \mathrm{~kg}\)
\(3 \mathrm{~kg}\)
Solution
Given, \(m_1=1 \mathrm{~kg}, m_2=2 \mathrm{~kg}\),
\(v_1=12 \mathrm{~ms}^{-1}, v_2=8 \mathrm{~ms}^{-1} \text { and } v_3=4 \mathrm{~ms}^{-1}\)
Since, in the explosion of stationary rock, the momentum is conserved,
so
\(\begin{aligned}
\mathbf{p}_i & =\mathbf{p}_f \\
\mathbf{0} & =\mathbf{p}_f=\mathbf{p}_1+\mathbf{p}_2+\mathbf{p}_3
\end{aligned}\)
where, \(\mathbf{p}_1=m_1 \mathbf{v}_1, \mathbf{p}_2=m_2 \mathbf{v}_2\) and \(\mathbf{p}_3=m_3 \mathbf{v}_3\)
\(\begin{aligned}
\mathbf{p}_3 & =-\left(\mathbf{p}_1+\mathbf{p}_2\right) \\
\mathbf{p}_3 & =\sqrt{\mathbf{p}_1^2+\mathbf{p}_2^2+2 \mathbf{p}_1 \cdot \mathbf{p}_2 \cos \theta} \\
\mathbf{p}_3 & =\sqrt{12^2+16^2+2 \times 12 \times 16 \cos 90^{\circ}} \\
\mathbf{p}_3 & =20 \\
m_3 v_3 & =m_3 \times 4=20 \Rightarrow m_3=5 \mathrm{~kg}
\end{aligned}\)
Hence, the mass of the rock is,
\(\begin{aligned}
& m=m_1+m_2+m_3 \\
& m=1+2+5=8 \mathrm{~kg}
\end{aligned}\)
So, the correct option is (a).