A homogeneous bar of length L can rotate on a horizontalfrictionless plane aroun

Question

A homogeneous bar of length L can rotate on a horizontalfrictionless plane
around a vertical axis passing through one of its ends (moment ofinertia I =
ML2/3). It is at rest until a small bullet of mass m with velocityv perpendicular to
the bar hits it in the mid point, bouncing back with velocityv’ = -v /3. Given that
the mass M of the bar is 9 times larger than the mass m of thebullet, calculate the
final kinetic energies of the bullet and of the bar. Determine ifthe collision was
elastic or not.

Explanation / Answer

as the torque of the impulse during the collision is 0 we conserve angular momentum we have mvl/2 =initial=-mvl/6+9ml*l*/3=final so =0.22v/l K.E of bullet=0.5*m*v*v/9=0.5*m*v*v/9 k.e of rod is0.5*9m*l*l*0.222v2/3l2=0.5*m*v*v*0.1452 if the collision were elastic then the fial energy =initialenergy=0.5*m*v*v the collision is not elastic

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