Hockey puck A initially travels at v 0 = 4.15 m/s in the direction shown on smoo

Question

Hockey puck A initially travels at  v0 = 4.15 m/s in the direction shown on smooth horizontal ice before striking hockey puck B which is initially at rest, off-center. After the collision, A is deflected at ? = 33.7 degrees 'above' its original direction, while puck B is deflected at ? = 30.8 degrees 'below' the original direction of puck A as shown. The pucks have the same mass.

a) What is the speed of puck A after the collision? (in m/s)

b) Referring to the same situation with v0 = 4.15, ? = 33.7 degrees, ?= 30.8 degrees, what percentage of the initial kinetic energy of the system is converted to other forms (i.e. dissipated) during the collision?

(Note: you will need to determine the final speed of the other puck B first.)

Explanation / Answer

Assume final speed of A is V and speed of B is W

1) momentum is conserved in both the horizontal and vertical direction

initially momentum in vertical direction = 0

final momentum in vertical direction = mAVsin(33.7)- mBWsin(30.8)=0 =>  mAV= 0.92*mBW--------(1)

initially momentum in horizontal direction =  mA*Vo= 4.15*mA

final momentum in horizontal direction = mAVcos(33.7) + mBWcos(30.8) = 4.15*mA------------------(2)

solving both the equations , V = 2.35 m/s and mBW= 2.55 mA vb = 2.55 m/s... va = 2.35 m/s

and both have same mass therefore W=2.55 m/s

a) the speed of the punk after the colission = V = 2.35 m/s

b) intial kinetic energy  =0.5*mA*4.152= 8.61*mA

final kinetic energy = 0.5*mA*2.352 +0.5*mB*2.552= 6.01*mA

percentage loss in kinetic energy = (8.61-6.01)*100/8.61 = 30.19%

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