Published by:
CGP EDU Academic Team
Published on: August 21, 2026
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
v
B
v
C
2 
v
v
D
3 
v
v
Text Solution
Verified by ExpertsThe correct answer is:
Given that,
Initial momentum = 0
Let the masses of 3 fragment be 2m, 2m, and m
Momentum along x-direction = 2mv 
Momentum along y-direction = 2mv 
Net momentum Now, 
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