Hockey puck B rests on a smooth ice surface and is struck by a second puck A, wh

Question

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially traveling at 16.0 and is deflected 24.0 from its initial direction. Assume that the collision is perfectly elastic. Part A Find the final speed of the puck B after the collision. SubmitMy AnswersGive Up Part B Find the final speed of the puck A after the collision. SubmitMy AnswersGive Up Part C Find the direction of B's velocity after the collision. from the initial A's direction SubmitMy AnswersGive Up Provide FeedbackContinue

Explanation / Answer

Since the masses are the same and the collision is elastic the angle between the pucks after the collision will be 90 degrees. So, assuming that the first puck is deflected downwards at a 25 degree angle, the second puck will be deflected upwards at a 65 degree angle to the horizontal. Conservation of momentum can be applied in the vetical and horizontal directions: call V the velocity of the first puck after the collision call U the velocity of the second puck after the collision call m the mass of each puck Vertical: before the collision this is just 0 after the collision the sum must also be 0 so mVsin(25) = mUsin(65) so: mVsin(25) = mUsin(65) V = Usin(65)/sin(25) Horizontal: before the momentum is m(15) after will be mVcos(25) + mUcos(65) so: m(15) = mVcos(25) + mUcos(65) 15 = Vcos(25) + Ucos(65) 15 = [Usin(65)/sin(25)]cos(25) + Ucos(65) 15 = 1.94358U + 0.42262U 15 = 2.36620U U = 6.3393 m/s V = Usin(65)/sin(25) = (6.33928)(2.14451) V = 13.5946 m/s check vertical: 6.3393sin(65) = 13.5946sin(25) 5.75 = 5.75 .... good check horizontal: 15 = 6.3393cos(65) + 13.5936cos(25) 15 = 2.68 + 12.32 = 15 .... good

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