Skater A, with mass 55.0 kg and initial speed 12.0 m/s, bumps into stationary Sk

Question

Skater A, with mass 55.0 kg and initial speed 12.0 m/s, bumps into stationary Skater B, mass 70.0 kg. After the collision, Skater A moves at an angle of 32.0degree from her original direction, while Skater B moves at an angle of 46.0degree from Skater A's original direction. Both skaters move on the frictionless, horizontal surface of the rink. What are the final speeds of Skater A and Skater B? Is kinetic energy conserved? If not, what is the change in total kinetic energy due to the collision?

Explanation / Answer

using momentum conservation,

initial momentum of system = final momentum of system

55 x 12i   +   70 x 0    = 55vA ( cos32i + sin32j) +   70vB ( cos46i - sin46j)

so along j vector,

55vAsin32 - 70vBsin46 = 0

vB = 0.58vA

along i vector

55vAcos32 + 70vBcos46 = 55x12

46.64vA + 48.63vA0.58 = 660

vA = 8.82 m/s

vB = 0.58 x 8.82 =5.11 m/s

b) initial KE = 55 x 12^2 / 2 = 3960 J

KEf = (55 x 8.82^2 / 2) + (70 x 5.11^2 /2) = 3053.21

no KE is not conserved.

loss = KEi - KEf = 906.79 J

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