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Unformatted text preview: homework 12 FRENNEA, KYLE Due: Apr 18 2007, 4:00 am 1 Question 1 part 1 of 3 10 points A block of mass 0 . 13 kg is attached to a spring of spring constant 23 N / m on a fric- tionless track. The block moves in simple har- monic motion with amplitude 0 . 18 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately be- fore it strikes the block is 57 m / s and the mass of the bullet is 2 . 6 g. 23 N / m . 13 kg 2 . 6 g 57 m / s Find the speed of the block immediately before the collision. Correct answer: 2 . 39422 m / s (tolerance 1 %). Explanation: Let : M = 0 . 13 kg , m = 2 . 6 g , A = 0 . 18 m , and k = 23 N / m . Since the block moves in simple harmonic motion, we know that x = A sin( t + ) and since the only force acting on it is the spring force F s =- k x , the equation of mo- tion F s = M a is M a = M d 2 x dt 2 =- k x , so the angular velocity of this initial motion is = radicalbigg k M = 13 . 3012 rad / s . The velocity at a given instant is found by differentiation v = d x dt = A cos( t + ) . The equilibrium position has the highest ve- locity, and the magnitude of the cosine func- tion is never greater than 1.0, so v eq = A = 2 . 39422 m / s . Question 2 part 2 of 3 10 points If the simple harmonic motion after the collision is described by x = B sin( t + ), what is the new amplitude B ? Correct answer: 0 . 263088 m (tolerance 1 %). Explanation: Convert the mass of the bullet to kg. Then use conservation of momentum M v eq + mv = ( M + m ) v f , since the bullet stays inside the block after collision. Both terms on the left have the same sign since the block was going from left to right. Solving for v f , we obtain v f = M v eq + mv M + m = (0 . 13 kg) (2 . 39422 m / s) (0 . 13 kg) + (0 . 0026 kg) + (0 . 0026 kg) (57 m / s) (0 . 13 kg) + (0 . 0026 kg) = 3 . 46493 m / s . The system is in harmonic oscillation after the collision as well, but with a different frequency and amplitude. The new frequency is, using the new (total) mass, is f = radicalbigg k M + m = radicalBigg (23 N / m) (0 . 13 kg) + (0 . 0026 kg) = 13 . 1702 rad / s , homework 12 FRENNEA, KYLE Due: Apr 18 2007, 4:00 am 2 and the velocity is now v = B f at equilib- rium. But this velocity is just the velocity immediately after the collision, since it oc- curred at equilibrium, so v f = B f , solving for B B = radicalbigg M + m k v f = radicalBigg (0 . 13 kg) + (0 . 0026 kg) (23 N / m) (3 . 46493 m / s) = 0 . 263088 m . Alternative Solution: An equivalent ap- proach (once we have v f ) is conservation of energy, using the fact that the potential en- ergy of the spring once it is fully compressed (maximum amplitude) is E p = (1 / 2) k B 2 , or 1 2 ( M + m ) v 2 f = 1 2 k B 2 , so by solving for B we arrive at the same result....
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