Two ice skaters, Daniel (mass 69.7 kg) and Rebecca (mass 50.4 kg) are practicing

Question

Two ice skaters, Daniel (mass 69.7 kg) and Rebecca (mass 50.4 kg) are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 12.6 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 6.6 m/s at an angle of 52.5° from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

(a) What are the magnitude and direction of Daniel's velocity after the collision?
  m/s ?
angle ? (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?
  J ?

Explanation / Answer

Linear momentum is conserved.
Consider R's initial direction to be the x-axis.
vR1x = 12.6 m/s
vR2x = (6..6 m/s) cos(52.5 deg) = 4.017 m/s
vR2y = (6.6 m/s) sin(52.5 deg) = 5.236 m/s
Momentum components are
pR1x = (12.6 m/s)(50.4 kg) = 635.04 kg m/s
pR2x = (4.017 m/s)(50.4 kg) = 202.45 kg m/s
pR2y = (5.236 m/s)(50.4 kg) = 263.89 kg m/s

Thus, the components of Daniel's momentum after the collision are
pD2x = (635.04 - 202.45) kg m/s = 432.59 kg m/s
pD2y = -263.89kg/ms
The components of Daniel's velocity after the collision are found by dividing out his 69.7-kg mass:
vD2x = 6.206 m/s
vD2y = -3.786 m/s

a) Magnitude of vD2 = sqrt(6.206^2 + 3.786^2)
= 7.26 m/s
Direction of vD2 = arctan(-3.786/6.206)
= -31.38 degrees, i.e., 89.8 degrees from Rebecca's new direction, or 31.38 deg from her old direction (but turned in the opposite sense from the way she turned)

b) Kinetic energy: the 90-degree separation suggests that almost no kinetic energy was lost, but you should figure out (1/2)mv^2 for each skater after the collision, and subtract their sum from the original KE of (1/2)(50.4 kg)(12.6)^2 = 4000.75 J.

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