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 33.0 from its original direction. (See (Figure 1) .) Puck B acquires a velocity at a 44.5 angle to the original direction of A. The pucks have the same mass.

a) Compute the speed of puck A after the collision. Express your answer in meters per second to three significant figures.

b) Compute the speed of puck B after the collision. Express your answer in meters per second to three significant figures.

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

For initial velocities let us use the symbol u
For final velocities we use v.
For the puck A uA means initial velocity of A and so on
--------------------------------------...

In the y direction its initial momentum of A is zero.
but its component of final momentum in y direction is m vA sin 30.5
The component of final momentum of B in the y direction is – vB sin 42.5
Hence vA sin 33 -vB sin 44.5 = 0
have vA/vB = sin 44.5 / sin 33 = 1.29
vA =1.29vB.-------------1
-----------------------------
In the x direction, the initial momentum of A was uA = 40m m/s
Its component of final momentum is m vA sin 33
The component of final momentum of B in the x direction is mvB cos 44.5
40 = vA sin 33 + vB cos 44.5
From 1
40 = 1.29vB * cos 33 + vB cos 44.5
40 = 2 vB
vB = 20 m/s

vA = 1.29vB
vA == 25.8. m/s

============================
K.E of A after collision 0.5mvA^2
K.E of A before collision 0.5mu^2

Their ration = (vA/u) ^2= (25.8/40)^2 = 0.416 = 41.6% is the energy remaining after collision

Energy lost is 1-0.416=0.584 = 58.4%
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Consider the y direction .
Total momentum in this direction initially is zero.

Since masses are equal ,I have omitted in all equations
Final momentum is vA sin 1 -vB sin 2. and this must be zero, since total momentum remains the same before and after collision .
vA sin 1 -vB sin 2 = 0
We have vA/vB = sin 2 / sin 1 = sin 44.5 / sin 33 = 1.29
vA = 1.29vB -----------------1
-------------------------------------

In the x direction
uA. = vA cos 33 + vB cos 44.5
From 1
40 = 1.29vB * cos 33 + vB cos 44.5
40 = 1.8 vB
vB = 22.22 m/s
vA = 1.29*21.21 = 28.66. m/s
======================2

K.E of A before collision 0.5mv^2 = 0.5*m

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