Hockey puck A initially travels at v0 = 6.79 m/s in the direction shown on smoot

Question

Hockey puck A initially travels at v0 = 6.79 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 = 33.1 degrees 'above' its original direction, while puck B is deflected at = 24.2 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 = 6.79, ? = 33.1 degrees, ?= 24.2 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(33.1)- mBWsin(24.2)=0 =>  mAV= 0.75*mBW--------(1)

initially momentum in horizontal direction =  mA*Vo= 6.79*mA

final momentum in horizontal direction = mAVcos(33.1) + mBWcos(24.2) = 6.79*mA------------------(2)

solving both the equations , V = 3.3 m/s and mBW= 4.4 mA

and both have same mass therefore W=4.4 m/s

a) the speed of the punk after the colission = V = 3.3 m/s

b) intial kinetic energy  =0.5*mA*6.792= 23.05*mA

final kinetic energy = 0.5*mA*3.32 +0.5*mB*4.42= 5.45*mA + 9.68*mA= 15.13*mA

percentage loss in kinetic energy = (23.05-15.13)*100/23.05 = 34.36

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