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 :

Question

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) 3 √2 v (B) v (C) √2 v (D) 2 √2 v

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b6b9e1c397726ff89a0f15efcbec3a28-.png" /> By conservation of momentum: m(0)=(2 m/5)(-v hati)+(2 m/5)(-v hatj)+(m/5) vecv prime ⇒ vecv prime=2 v hati+2 v hatj ⇒ v prime=√(2 v)2+(2 v)2 =2 √2 v

Q. 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 :

$32 v$

$v$

$2 v$

$22 v$

By conservation of momentum :
$m (0) = \frac{2 m}{5} (- v \hat{i}) + \frac{2 m}{5} (- v \hat{j}) + \frac{m}{5} v^{'}$
$\Rightarrow v^{'} = 2 v \hat{i} + 2 v \hat{j}$
$\Rightarrow v^{'} = (2 v)^{2} + (2 v)^{2}$
$= 22 v$

Q. 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 :

32​v

v

2​v

22​v

By conservation of momentum : m(0)=52m​(−vi^)+52m​(−vj^​)+5m​v′ ⇒v′=2vi^+2vj^​ ⇒v′=(2v)2+(2v)2​ =22​v

Q. 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 :

$32 v$

$v$

$2 v$

$22 v$

By conservation of momentum :
$m (0) = \frac{2 m}{5} (- v \hat{i}) + \frac{2 m}{5} (- v \hat{j}) + \frac{m}{5} v^{'}$
$\Rightarrow v^{'} = 2 v \hat{i} + 2 v \hat{j}$
$\Rightarrow v^{'} = (2 v)^{2} + (2 v)^{2}$
$= 22 v$