-3 points YF14 8.P027 Two ice skaters, Daniel (mass 69.4 kg) and Rebecca (mass m

Question

-3 points YF14 8.P027 Two ice skaters, Daniel (mass 69.4 kg) and Rebecca (mass moving at 13.3 m/s before she collides with him. After the collision, Rebecca has a velocity of skaters move on the frictionless, horizontal surface of the rink. 50.3 kg) are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, wheis magnitude 6.7 m/'s at an angle of 53.9 from her intial direction. Both (a) What are the magnitude and direction of Daniel's velocity after the collision? m/s (clockwise from Rebecca's original direction of motion) (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Explanation / Answer

Given,

mD = 69.4 kg ; mR = 50.3 kg ; uR = 13.3 m/s ; vR = 6.7 m/s at theta = 53.9 deg

A)momentum is conserved.

from conservation of momentum in X direction

mR uR = mR vR cos53.9 + mD vD cos(theta1)

50.3 x 13 = 50.3 x 6.7 x 0.59 + 69.4 vD cos(theta1)

vD cos(theta1) = 6.77 (1)

from conservation of momentum in Y direction

0 = mR vR sin(theta) - mD vD sin(theta1)

0 = 50.3 x 6.7 x sin53.9 - 69.4 vD sin(theta1)

vD sin(theta1) = 3.92 (2)

(2)/(1)

tan(theta1) = 3.92/6.77 = 0.579

theta1 = tan^-1(0.579) = 30.07 deg

vD = 6.77/cos(theta1) = 6.77/cos30.07 = 7.82 m/s

Hence, vD = 7.82 m/s

theta1 = 30.07 deg

c)KEi = 1/2 mR uR^2

KEi = 0.5 x 50.3 x 13.3^2 = 4448.78 J

Kef = 1/2 mR vR^2 + 1/2 mD vD^2

KEf = 0.5 (50.3 x 6.7^2 + 69.4 x 7.82^2) = 3250.97 J

change = KEf - KEi

change = 3250.97 - 4448.78 = -1197.81

Hence, change = -1197.81 J

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