First slide

The steps in collision

Question

A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular directions each with speed v. The energy released during the process is [NEET (Odisha) 2019]

Moderate

A

$\frac{3}{5} {mv}^{2}$
B

$\frac{5}{3} {mv}^{2}$
C

$\frac{3}{2} {mv}^{2}$
D

$\frac{4}{3} {mv}^{2}$

Solution

The particle of mass 5m breaks into three fragments of masses m, m and 3m, respectively. Two fragments of
mass m each move in perpendicular directions with velocity v and the left fragment will move in a direction with velocity v' such that the total momentum of the system must remain conserved.
By law of conservation of momentum,
$5 m \times 0 = m v \hat{i} + m v \hat{j} + 3 m v'$
$\Rightarrow$ $v' = - \frac{v}{3} \hat{i} - \frac{v}{3} \hat{j}$
$∴$ $|v^{'}| = \sqrt{{(- \frac{v}{3})}^{2} + {(- \frac{v}{3})}^{2}} = \frac{v \sqrt{2}}{3}$
$∴$ Energy released,
$E = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} \times 3 m {(\frac{v \sqrt{2}}{3})}^{2}$
$= m v^{2} + \frac{m v^{2}}{3} = \frac{4}{3} m v^{2}$

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