A body of mass $$5$$ kg explodes at rest into three fragments with masses in the ratio $$1$$ : $$1$$ : $$3$$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $$21$$ $$m/s$$. The velocity of the heaviest fragment will be

Question

Infinity Learn · Accepted Answer

$$Px=m$$×$$vx=1$$×$$21=21$$ kg m​$$/$$​$$sPy=m$$×$$vy=1$$×$$21=21$$ kg m​$$/$$​s∴ Resultant $$=$$ $$Px2+Py2=212$$ kg $$m/sThe$$ momentum of heavier fragment should be numerically equal to resultant of P→x and P→y. $$3$$×$$v=Px2+Py2=212$$   ∴$$v=72$$ $$=$$ $$9.89$$ $$m/s$$

A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be

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