A bullet of mass mB = 0.01 kg is moving with a speed of 132 m/s when it collides
ID: 2197130 • Letter: A
Question
A bullet of mass mB = 0.01 kg is moving with a speed of 132 m/s when it collides with a rod of mass mR = 9 kg and length L = 5 m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating.
(a) Find the angular velocity, ?, of the bullet?rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
(b) How much kinetic energy is lost in the collision?
Explanation / Answer
Initially, L B (angular momentum of bullet about the hinge) = mB*v*perpendicular distance = (mB)vL/4. Finally, L S (L system) = I(system)?. [ I = moment of inertia] I system = mR * L² /12 + mB * (L/4)² [assuming the rod to be uniform, its COM is at its centre. I about that point = m l² /12) = mR * L² /12 + mB * L² / 16 = L² [mR/12 + mB/16]. Conserving L, (mB) v L / 4 = L² [mR/12 + mB/16]? => mB v = L (mR / 3 + mB /4) ? => ? = mB v / L (mR / 3 + mB /4) . Putting values, ? = 0.0127*101/1.09(7.23/3 + 0.0127/4) rad/s = 1.2827/2.63 = 0.488 rad/s (Answer). K Initial = 1/2 mB v² = 0.5*0.0127*)(101)² = 64.77635 Joule. K Final = 1/2 I ?² = 0.5 * 0.6 * (1.09)² * (0.488)² = 0.17 Joule. So, Energy lost = 64.6 Joule (Answer)
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