Q.

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,

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a

43mv2

b

35mv2

c

53mv2

d

32mv2

answer is A.

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Detailed Solution

From conservation of linear momentum,

Question Image

0=mvj^+mvi^+3mv1 v1=-v3(i^+j^) v1=23v

From conservation of energy,

KEi=0  ; KEf=12mv2+12mv2+12(3 m)232v2 =mv2+mv23=43mv2   ;    ΔKE=KEf-KEi=43mv2

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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,