A hockey puck B rests on a smooth surface of ice and is struck by a second puck

Question

A hockey puck B rests on a smooth surface of ice and is struck by a second puck A , which was originally traveling at 40.0 m/s and which is deflected 31.5 from its original direction. (See (Figure 1) .) Puck B acquires a velocity at a 43.0 angle to the original direction of A . The pucks have the same mass.

Part A

Compute the speed of puck A after the collision.

Express your answer in meters per second to three significant figures.

Part B

Compute the speed of puck B after the collision.

Express your answer in meters per second to three significant figures.

Part C

What fraction of the original kinetic energy of puck A dissipates during the collision?

Express your answer as a percentage to three significant figures.

Explanation / Answer

mA = = mB , uA = 40m/s, uB =0 ,

vAx = vAcos(31.5), vAy =vA sin(31.5)

vBx = vB cos(43) , vBy = - vBsin(43)

From conservation of momentum along x direction

mAuAx +mB uBx = mAvAx +mBvBx

40 = vAcos(31.5) +vB cos(43) .. (1)

From conservation of momentum along y direction

mAuAy +mB uBy = mAvAy +mBvBy

0 = vAsin(31.5) - vB sin(43) .. (2)

From (1) and (2) we get

(a) vA = 28.3 m/s

(b) vB = 21.7 m/s

KAf/KAi = (1/2)mvA^2/(1/2)muA^2 = vA^2/uA^2 =(28.3/40)^2 =0.5 J

Energy lost is = 1-0.5 =0.5 =50%

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