Two ice skaters, Daniel (mass 70.0 kg ) and Rebecca (mass 45.0 kg ), are practic

Question

Two ice skaters, Daniel (mass 70.0 kg ) and Rebecca (mass 45.0 kg ), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 55.1 from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.

Part A What is the magnitude of Daniel's velocity after the collision? v = m/s

Part B What is the direction of Daniel's velocity after the collision? = from the Rebecca's original direction

Part C What is the change in total kinetic energy of the two skaters as a result of the collision? K =

Explanation / Answer

Assuming rebecca was moving in +x direction initially

Mass of Danield m1 = 70 Kg

Mass of Rebecca m2 = 45 Kg

Initial velocity of daniel u1 = 0m/s

Initial velocity of rebecca u2 = 13 m/s

we need to find final velocity of daniel

Applying conservation of momentum in x axix we have:

m1u1 + m2u2 = m1vcos(theta) + m2v2cos(55.1)

70*0 + 45*13 = 70vcos(theta) + 45*8*0.57

vcos(theta) = 5.43 ----- (1)

Applying conservation of momentum in y axis

0 = -m1vsin(theta)+ m2v2sin(55.1)

70*vsin(theta) = 45*8*0.82 = 295.2

vsin(theta) = 295.2/70 = 4.22 -----(2)

squaring and adding 1 and 2 we have

v^2 = 5.43*5.43 + 4.22*4.22

part a) v = 6.88 m/s

dividing 2 by one we have

tan(theta) = 4.22/5.43

part b) theta = 37.95 degrees

part c)

change in K.E. = initial - final K.E. = (0.5*45*13*13) - (0.5*45*8*8) - (0.5*70*6.88*6.88) = 705.796 J

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