A hockey puck B resting on a smooth ice surface is struck by a second puck A whi

Question

A hockey puck B resting on a smooth ice surface is struck by a second puck A which was originally traveling at 50 m/s. After colliding, the speed of puck A is reduced to 35 m/s and is deflected through an angle from its original direction. Puck B acquires a velocity of 25 m/s at an angle with the direction of original velocity of puck A. Assume that both pucks A & B have the same mass and that the collision is not perfectly elastic. (See the Figure.) (a) Find the angles and . (b) What fraction of the original K.E. of A is lost?

Explanation / Answer

mA = mB

uA = 50 m/s, uB = 0 , vA = 35 m/s , vB = 25 m/s

From conservation of mometum along X direction

mAuAx +mBuBx = mAvAx +mBvBx

50 +0 = 35cos () + 25 cos() ...(1)

From conservation of mometum along Y direction

mAuAy +mBuBy = mAvAy +mBvBy

0 = 35sin() +25 sin () ...(2)

= 27.66 degrees above the original direction of puck A

= 40.54 degrees below above the original direction of puck A

Part (b) :

(K1 -K2)/K1 = ((1/2)muA^2 -( (1/2)mvA^2 +(1/2)mvB^2))/(1/2)muA^2

=( uA^2 - vA^2 -vB^2)/uA^2

= ((50*50) - (35*35) -(25*25))/(50*50) =0.26

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