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 14.0 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. 1.What is the magnitude of Daniel's velocity after the collision? 2.What is the direction of Daniel's velocity after the collision? 3What is the change in total kinetic energy of the two skaters as a result of the collision?

Explanation / Answer

m1(Daniel) = 70 kg              m2(Rebeca) = 45 kg

before collision

speeds

u1x = 0                       u2x = 14

u1y = 0                          u2y = 0

after collision

v1x =                                   v2x = 8*cos55.1

v1y = ?                              v2y = 8*sin55.1

from momentum conservation

along x axis

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

70*0 + 45*14 = (70*v1x)+ (45*8*cos55.1)

v1x = 6.1 m/s <<-------answer

along y

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

0 + 45*0 = (70*v1y)+(45*8*sin55.1)

v1y = -4.22 m/s <<------answer

1)

magnitude = v1 = sqrt(v1x^2+v1y^2)

v1 = sqrt(6.1^2+4.22^2) = 7.42 m/s

+++++++++++++++++++

2)

direction

tan^-1(v1y/v1x) = 34.7 degrees

_______________

3)

b)

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEi = (0.5*70*0^2)+(0.5*45*14^2) = 4410 J

KEf = 0.5*m1*v1^2 + 0.5*m2*v2^2

KEf = (0.5*70*7.42^2)+(0.5*45*8^2) = 3366.974 J

change = 4410-3366.974 = 1043.026 J

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