Two ice skaters, Daniel (mass 65.0 kg ) and Rebecca (mass 45.0 kg ), are practic

Question

Two ice skaters, Daniel (mass 65.0 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 6.00 m/s at an angle of 52.1 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?

Part B

What is the direction of Daniel's velocity after the collision?

Part C

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

Part A

What is the magnitude of Daniel's velocity after the collision?

Explanation / Answer

Given that m1 = 45 kg and u1 = 14 m/s

m2 = 65 kg and u2 = 0 m/s

using law of conservation of momentum along original dorection

m1*u1 = (m1*v1*cos(52.1)) + (m2*v2*cos(theta))

45*14 = (45*6*cos(52.1))+(65*v2*cos(theta))

v2*cos(theta) = 7.14 m/s

along perpendicular to the original direction

0 = 45*6*sin(52.1) + (65*v2*sin(theta))

v2*sin(theta) = -3.27 m/s

then v2*cos(theta) / v2*sin(theta) = -7.14/3.27 = -2.18

cot(theta) = -2.18

theta = cot^(-1)(-2.18)

theta = -24.64 degrees

i.e 24.64 degrees below the original direction

v2 = 7.14/cos(-24.64) = 7.85 m/s is the answer for part A)

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Part B

direction is 24.64 degrees below the original direction of rebecca

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Part C)

Final KE = 0.5*m1*v1^2 + 0.5*m2*v2^2 = (0.5*45*6^2)+(0.5*65*7.85^2) = 2812.73 J

Initial KE = 0.5*m1*u1^2 = 0.5*45*14^2 = 4410 J

change in total KE = 2812.73 - 4410 = -1597.27 J

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