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Q. Two identical spheres P and Q lie on a smooth horizontal circular groove at opposite ends of a diameter. P is projected along the groove and at the end of time t , impinges on Q . If e is the coefficient of restitution, then the second impact will occur after the shortest time of

NTA AbhyasNTA Abhyas 2022

Solution:

Let r be the radius of the groove.
As P and Q lie at opposite ends of a diameter, therefore distance covered =12(2πr)=πr
SpeedofPbeforeimpact,u=πrt
Speed of Q before impact =0 .
Let v and v be velocities of P and Q respectively after impact.
vv=e(u0)=eu
Thus the velocity of Q relative to P after the first impact = eu
Let the next impact occur after t seconds. This would happen when Q completes one round more than P in time t .
Distance travelled in one round = 2πr
t=Relative distanceRelative velocity=2πrvv
t=2πreu=2(ut)eu=2te