SATHEE: Centre of Mass 3 Question 27

Centre of Mass 3 Question 27

30. A particle of mass $4 m$ which is at rest explodes into three fragments. Two of the fragments each of mass $m$ are found to move with a speed $v$ each in mutually perpendicular directions. The total energy released in the process of explosion is

$(1987, 2 M)$

Integer Answer Type Questions

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Solution:

From conservation of linear momentum $p_{3}$ should be $\sqrt{2} m v$ in a direction opposite to $p_{12}$ (resultant of $p_{1}$ and $p_{2}$ ). Let $v^{'}$ be the speed of third fragment, then

$\begin{aligned} (2 m) v^{'} & = \sqrt{2} m v \\ ∴ v^{'} & = \frac{v}{\sqrt{2}} \end{aligned}$

$∴$ Total energy released is,

$\begin{aligned} E & = 2 \frac{1}{2} m v^{2} + \frac{1}{2} (2 m) v^{' 2} \\ = m v^{2} + m {\frac{v}{\sqrt{2}}}^{2} = \frac{3}{2} m v^{2} \end{aligned}$

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Centre of Mass 3 Question 27

30. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is

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30. A particle of mass $4 m$ which is at rest explodes into three fragments. Two of the fragments each of mass $m$ are found to move with a speed $v$ each in mutually perpendicular directions. The total energy released in the process of explosion is