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Unformatted text preview: Midterm Solutions (I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel (a solid disc) of mass M , radius R , anchored at its center but free to rotate. (i) Which of energy, momentum and angular momentum is conserved for the bullet+wheel system? Give a few words of explanation. (ii) Find f the final angular velocity of the wheel. 10 (i) Only angular momentum is conserved. The collision is an inelastic collision, so energy cannot be conserved. Momentum is not conserved because there is an external force acting at the center of the wheel that keeps the wheel-bullet system from moving forward after the collision. Angular momentum is conserved because there are no external torques acting on the system. (The force acting at the center of the wheel does not provide a torque because it is acting at the pivot point.) (ii) To find f , we equate the angular momentum of the system just before the bullet hits the wheel with the final angular momentum of the rotating bullet-wheel system. mvR = f ( I wheel + I bullet ) = f 1 2 MR 2 + mR 2 f = v R 1 1 + M 2 m ! . Please note that linear momentum p and angular momentum L are not equivalent. They have separate conservations laws, different units, and measure fundamentally different quantities. You cannot equate p to L . (II) A block of mass M sits on frictionless table L meters from the edge. At t = 0 bullet of mass m and velocity v 1 penetrates it from left and exits to the right with a speed v 2 . (i) When will the block fly off the table? (ii) If the table has a height h how far from the edge of the table will it land? 10 (Neglect loss of wood in block due to bullet penetrating it and the time it takes bullet to traverse block.) (i) To find when the block will fly off the table, we need to calculate the final velocity of the block by conserving momentum before and after the collision. mv 1 = mv 2 + Mv = v = m M ( v 1- v 2 ) There are no external forces on the block, so the block will fly off the table at t = L v = M m L v 1- v 2 . (ii) To find the distance the block travels in the x direction after leaving the table, we first need to find the total time it spends in the air. We know that the block falls a distance h in the y direction and has no initial velocity in y . This tells us that h = 1 2 gt 2 = t = s 2 h g (clearly, the negative solution for t is not physical for this problem). There are no forces in the x direction, so the total distance the block travels once it leaves the table is x = vt = ( v 1- v 2 ) m M s 2 h g ....
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