NEET 2022 — Physics solution

NEET2022Physics

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is

A
$2 v$
B
$22 v$
C
$32 v$
D
$v$

Solution

The fragment masses are $\frac{2 m}{5}$ , $\frac{2 m}{5}$ , and $\frac{m}{5}$ . Take the two equal fragments to move along $- \hat{i}$ and $- \hat{j}$ with speed $v$ .

Initial momentum is zero, so by conservation of momentum:

$\frac{2 m}{5} (- v \hat{i}) + \frac{2 m}{5} (- v \hat{j}) + \frac{m}{5} v^{'} = 0$ .

Solving gives $v^{'} = 2 v \hat{i} + 2 v \hat{j}$ , so $∣ v^{'} ∣ = (2 v)^{2} + (2 v)^{2} = 22 v$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →