For rotational collisions we usually conserve angular momentum.
(i) But is linear momentum conserved as well?
(ii) If the coefficient of restitution is given, how do you calculate the question?
Ex. A stick of length L and mass M lies on a frictional horizontal surface on which it is free to move in any way. A ball of mass m moving with speed v collides with the stick at a distance d from the centre of stick with coefficient of restitution e=1. What must be the mass of the ball so that it (ball) remains at rest immediately after collision.
Soln: we can conserve linear momentum as mv=MV
but as COR = 1, V-0 = -1(0-v) i.e. v=V ==> m=M
BUT if we conserve angular momentum and then conserve kinetic energy(collision is elastic) then we get m =ML2/(L2 + 12d2)
Why is this
Further, if COR was not equal to 1, then we could not have conserved kinetic energy. Then how could we solve it?
33U have written
but as COR = 1, V-0 = -1(0-v) ...it is wrong..
velocity of separation of point of contact after collision=e. velocity of approach of point of contact..
U have taken velocity of COM... but it should be V+ωd..
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so V+ωd=v..
MV=mv..
and mvd=ML2/12ω
Or what U have done energy conservation.. is right as.. there by conserving energy u have applied condition of elastic collison also..
See the bold words..
106thanks abhisek
please solve Q37,52,43 from rotation probs too. the rest i have solved.