NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
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 :
- Av
- Bv
- C2v
- D3v
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Correct answer: C
Momentum of the system would remain conserved.
Initial momentum = 0
Final momentum should also be zero.
Let masses be 2m, 2m and m
Momentum along x-direction = 2mv$\widehat i$
Momentum along y-direction = 2mv$\widehat j$
Net momentum = $$\sqrt {{{(2mv)}^2} + {{(2mv)}^2}} = \sqrt 2 \,.\,2$$ mv
Now, 2$\sqrt2$mv = mv'
v' = 2$\sqrt2$v
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