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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.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=43mv2
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