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

As there is no diagram, assumming original motion to be parallel to x axis,
Let the mass be m.
Vai = 50 i^ m/s
Vbi = 0

Final Velocity,
Vaf = 35*cos() i^ + 35*sin() j^ (x1i^ + y1j^)
Vbf = 25*cos() i^ - 25*sin() j^ (x2i^ + y2j^)

Using Momentum Conservation,
50 = 35*cos() + 25*cos()
50 = x1 + x2 ---------1

35*sin() = 25*sin()
y1 = - y2 -------------2


|v1|^2 = x1^2 + y1^2 = 35^2 and |v2|^2 = x2^2 + y2^2 = 25^2, thus
35^2 - 25^2 = |v1|^2 - |v2|^2 = x1^2 - x2^2 = (x1 - x2) * (x1 + x2)
35^2 - 25^2 = (x1 - x2) * 50
x1 - x2 = (35^2 - 25^2) / 50
x1 - x2 = 12.
x1 + x2 = 50.

x1 = 31
x2 = 19

35*cos() = 31
= 27.66

25*cos() = 19
= 40.54

(b)
K.E lost = (K.Ein - K.Efin)/K.Ein * 100%
K.E lost = 1/2*m (vi^2 - (vf1^2 + vf2^2) )/ (1/2*m*vi^2) * 100%
K.E lost = [50^2 - (25^2 + 35^2)] / (50^2) * 100%
K.E lost = 26 %

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