
From momentum conservation, Pi0=Pf
mu=Mv…(i)
From angular momentum conservation about O,
mu⋅2L=12ML2ω
⇒ω=ML6mu…(ii)
From e=R.V⋅AR.V⋅S
1=uV+2ωL
V+2ωL=u
V+M3mu=u
Mmu+M3mu=u
M4mu=u
Mm=41
x=4
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses (Mm) is x1. The value of x will be
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