Questions
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 direction with speed v each. The energy released during the process is,
detailed solution
Correct option is A
From conservation of linear momentum.0=mvj^+mvi^+3mv→1 v→1=-v3(i^+j^) v1=23v KEi=0 KEf=12mv2+12mv2+12(3 m)232v2 =mv2+mv23=43mv2 ΔKE=KEf-KEi=43mv2Talk to our academic expert!
Similar Questions
In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 s is :
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