Source: winter 2008 midterm (modified to be a chapter 8, energy-type question) A

Question

Source: winter 2008 midterm (modified to be a chapter 8, energy-type question) A 15 kg block is on a 10 m long and 8 m high incline. It is attached via a light rope that passes over a light, pulley on a frictionless axis to a 5 kg block which is attached to the top of a spring with a spring constant of 24N/m that is initially neither stretched nor compressed. when released from rest , the 15kg block starts sliding down the incline. Find the common speed of the two blocks when the 15kg block has slid 0.5 m down the incline assuming the incline is frictionless. Repeat art a) assuming the coefficient of kinetic friction between the incline and the 15 kg block is mu = 0.20. A hockey puck B rests on a smooth ice surface and is struck by a second puck A. which was originally traveling at a 45.0 degree angle to the original direction of A as shown. The pucks have the same mass. Compute the speed of each puck after the collision. what fraction of the original kinetic energy of puck A dissipates during the collision?

Explanation / Answer

initial velocity of the puck A , vA = 40 m/s let initial speed of the puck B is vB = 0 m/s let final speed of the puck A is v'A let final speed of the puck B is v'B a) x -component : from law of conservation of momentum ,       mAvA + mBvB = m'Av'A cos30 + m'Bv'B cos45 here , mass mA = mB       vA + 0 = v'A cos30 + v'B cos45       (3 / 2)v'A + (1 / 2)v'B = vA   ................. (1) y -component : from law of conservation of momentum ,       0 + 0 = v'A sin30 - v'B sin45 here , mass mA = mB       (1 / 2)v'A - (1 / 2)v'B =  0   ................. (2) by solving eq (1) and (2) , we get        v'A = 29.27 m/s this value substitute in eq (2) , we get        v'B = 20.70 m/s ............................................................................. fraction of original kinetic energy of puck A dissipates during the collision is        = (1/2)[mvA2 - mv'A2] / (1/2)mvA2        = [vA2 - v'A2] / vA2        = [40 2 - 29.27 2] / 40 2        = 0.4643        = 46.43% a) x -component : from law of conservation of momentum ,       mAvA + mBvB = m'Av'A cos30 + m'Bv'B cos45 here , mass mA = mB       vA + 0 = v'A cos30 + v'B cos45       (3 / 2)v'A + (1 / 2)v'B = vA   ................. (1) y -component : from law of conservation of momentum ,       0 + 0 = v'A sin30 - v'B sin45 here , mass mA = mB       (1 / 2)v'A - (1 / 2)v'B =  0   ................. (2) by solving eq (1) and (2) , we get        v'A = 29.27 m/s this value substitute in eq (2) , we get        v'B = 20.70 m/s ............................................................................. fraction of original kinetic energy of puck A dissipates during the collision is        = (1/2)[mvA2 - mv'A2] / (1/2)mvA2        = [vA2 - v'A2] / vA2        = [40 2 - 29.27 2] / 40 2        = 0.4643        = 46.43%       (1 / 2)v'A - (1 / 2)v'B =  0   ................. (2) by solving eq (1) and (2) , we get        v'A = 29.27 m/s this value substitute in eq (2) , we get        v'B = 20.70 m/s ............................................................................. fraction of original kinetic energy of puck A dissipates during the collision is        = (1/2)[mvA2 - mv'A2] / (1/2)mvA2        = [vA2 - v'A2] / vA2        = [40 2 - 29.27 2] / 40 2        = 0.4643        = 46.43%        = 0.4643        = 46.43%

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