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Question

Two steel balls of equal diameter are connected by a rigid bar of negligible weight as shown and are dropped in the horizontal position from height h above the heavy steel and brass base plates. If the coefficient of restitution between the ball and the steel base is 0.6 and that between the other ball and the brass base is 0.4, the angular velocity of the bar immediately after rebound is (Assume the two impacts are simultaneous and take g=10 m/s2).


A
25 rad/s
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B
15 rad/s
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C
35 rad/s
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D
14 rad/s
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Solution

The correct option is A 25 rad/s
If v is the velocity with which the balls strike the bases, then their velocities after collision are 0.6v and 0.4v as shown.
Let u be the velocity of COM just after collision and ω is the angular velocity acquired by the rod. Since diameters of the steel balls are same, their masses are also same and hence COM will be at l2 from A


For end A: 0.6v=u+12ωl ...(i)
For end B: 0.4v=u12ωl ...(ii)
Subtracting (i) and (ii) we get
0.2v=ωlω=v5l

Now using [v2=u2+2as]
Velocity of balls before collsion, v is given by
v2=0+2×10×0.2=4
v=2 m/s
or ω=25×1=25 rad/sec

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