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 16.0m/s and is deflected 22.0 degrees from its initial direction. Assume that the collision is perfectly elastic.

a) Find the final speed of puck b after the collision.

b) Find the final speed of puck a after the collision.

c) Find the direction of b's velocity after the collision

Answer 1

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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 travelling at 15.0 m/s and is deflected 25.0degrees from its initial direction. assume that the collision is perfectly elastic. find the final speed of each puck and the direction of puck B's velocity after the collision. thank you.

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 ....