First slide

Location of centre on mass for multiple point masses

Question

A body of mass(4m) is lying in x–y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds ( $ν$ ). The total kinetic energy generated due to explosion is

Difficult

A

$m v^{2}$
B

$\frac{3}{2} m v^{2}$
C

$2 m v^{2}$
D

$4 m v^{2}$

Solution

Let ${\vec{v}}^{'}$ be velocity of third piece of mass 2 m. Initial momentum, ${\vec{p}}_{i} = 0$ (As the body is at rest)
Final momentum, ${\vec{p}}_{f} = m v \hat{ı} + m v \hat{ȷ} + 2 m {\vec{v}}^{'}$
$According to law of conservation of momentum$
$\begin{array}{l} {\vec{p}}_{i} = {\vec{p}}_{f} \\ 0 = m v \hat{i} + m v \hat{j} + 2 m {\vec{v}}^{'} \\ {\vec{v}}^{'} = - \frac{v}{2} \hat{i} - \frac{v}{2} \hat{j} \end{array}$
$The magnitude of {\vec{v}}^{'} is$
$v' = \sqrt{{(- \frac{v}{2})}^{2} + {(- \frac{v}{2})}^{2}} = \frac{v}{\sqrt{2}}$
$Total kinetic energy generated due to explosion$
$\begin{array}{l} = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{' 2} \\ = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} (2 m) {(\frac{v}{\sqrt{2}})}^{2} \\ = m v^{2} + \frac{m v^{2}}{2} = \frac{3}{2} m v^{2} \end{array}$

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