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

Initial momentum = 50*m
Final momentum of A in x-direction = m*35cos(alpha)
Final momentum of A in y-direction = m*35sin(alpha)
Final momentum of B in x-direction = m*25cos(beta)
Final momentum of B in y-direction = m*25sin(beta)

from conservation of momentum
50m = 35mcos(alpha) + 25mcos(beta)
35msin(alpha) = 25msin(beta)

from 2, sin(alpha) = 5sin(beta)/7
from 1, cos(alpha) = 5[2 - cos(beta)]/7

squaring and adding, 1 = 25[sin^2(beta) + 4 + cos^2 (beta) - 4cos(beta)]/49
49 = 25[5 - 4cos(beta)]
[5-49/25]/4 = cos(beta)
beta = 40.5358 deg
and sin(alpha) = 5sin(beta)/7
alpha = 27.660 deg

2. Original ke of A = 0.5m(50)^2
Final ke = 0.5m(35)^2
Fraction lost = 0.5m[50^2 - 35^2]/0.5m(50)^2 = 1 - (35/50)^2 = 0.51 = 51 pc

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