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 v = 40.0 m/s and which is deflected 1 = 27.0 from its original direction. (See the figure below (Figure 1) .) Puck B acquires a velocity at a 2 = 45.0 to the original direction of A . The pucks have the same mass.

Compute the speed of puck A after the collision

Compute the speed of puck B after the collision.

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

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 vA sin 27
The component of final momentum of B in the y direction is – vB sin 45
Hence vA sin 27 -vB sin 45 = 0
have vA/vB = sin 45 / sin 27 = 1.56
vA =1.56vB.
In the x direction, the initial momentum of A was uA = 40m m/s
Its component of final momentum is mvA sin 27
The component of final momentum of B in the x direction is mvB cos 45
40 = vA sin 27 + vB cos 45
From 1
40 = 1.56vB * cos 27 + vB cos 45
vB = 19.074 m/s

vA = 1.56vB
vA = 1.56 *19.074
vA == 29.755. m/s

Their ratio = (vA/u) ^2= (29.755/40)^2 = 0.55= 55% is the energy remaining after collision

Energy lost is 1-0.55=0.45 = 45%

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 45 / sin 27 = 1.56
vA = 1.56vB

In the x direction
uA. = vA cos 27 + vB cos 45
From 1
40 = 1.56vB * cos 27 + vB cos 45
40 = 2.1 vB
vB = 19.074m/s
vA = 2.1 * 19.74 = 40. m/s

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