Before CollisionAfter Collision A rod of mass M 2 kg, length L-1.8 meters, and m

Question

Before CollisionAfter Collision A rod of mass M 2 kg, length L-1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod is at rest when a puck of mass m0.2 kg approaches with a speed v, - 12 m/s perpendicular to the rod's length and strikes the rod at a with a speed Vf=6 with speed h angular speed w (a) What is the speed v of the rod's center of mass after the collision? m/s (b) what is the angular speed of the rod's rotation after the collision? rad/s (c) What is the change in kinetic energy of the system during the collision?

Explanation / Answer

As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.

So conserving linear momentum,

mvi+0=-mvf+MV....(final momentum of ball will be negative as it's direction is reversed)

0.2*12=(-0.2*6)+(2*V)

3.6=2V

V=1.8m/s

b)Now conserving the angular momentum after collision,

mvid=-mvfd+I

I=md(vi+vf)

(ML2/12)=18*0.3*0.2

(2*1.82/12)=1.08

=2 rad/s

c)

Now initial kinetic enrgy of the system

mv2/2

=0.2*122/2

=14.4J

Final kinetic enrgy of the system will be

mvf2/2+MV2/2+ML22/24

=7.92J

So, loss in enrgy will be

14.4-7.92

=6.48J

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