A bullet of mass m = 20 gm is traveling at a velocity of 200 m/s perpendicular t

Question

A bullet of mass m = 20 gm is traveling at a velocity of 200 m/s perpendicular to the long axis of a block of wood resting on a frictionless horizontal surface. The mass of the wood block is M = 10 kg, and its length L = 1m. The bullet bores through the very end of the block, emerging along its original path but with a reduced velocity of 100 m/s

(a) Determine the magnitude of the angular velocity of the block, assuming that the block is thin enough so the moment of intertia about its center of mass is I_cm = (1/12)ML^2

(b) Determine the linear speed of the center of mass of the block.

Explanation / Answer

moment of inertia =ML^2/12

applying law of conservation of angular momentum about the center of mass,

200*20*10^-3=10*(1/12)*w+20*10^-3*100

or w=2.4 rad/s

conserving linear momentum,

20*10^-3*200=10*v+20*10^-3*100

or v=0.2 m/s

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