Question

07.2 Skater A, with mass 80 kg and initial speed 9.0 m/s, runs into stationary Skater B, mass 65 kg. After the collision, Skater A moves at an angle of 55.0o from her original direction and Skater B moves at an angle of 10o from Skater A’s original direction. Both skaters move on the frictionless, horizontal surface of the rink. (a) What are the final speeds of Skater A and Skater B. (b) Is kinetic energy conserved? If not, what is the change in total kinetic energy due to the collision?

Answer #1

vAi = 9 i

vBi = 0

vAf = vA (cos55i + sin55 j)

vBf = vB (cos10i - sin10 j)

Applying momentum conservation,

mA vAi + mB vBi = mA vAf + mB vBf

720 i + 0 = (45.9 vA + 64 vB)i + (65.5 vA - 11.3 vB)j

65.5 vA - 11.3 vB = 0

vB = 5.8 vA

45.9 vA + 64 vB = 720

(puttin in vB in terms of vA)

vA = 1.73 m/s .....Ans

vB = 10 m/s ......Ans

(B) Ki = (80 x 9^2 )/2 = 3240 J

Kf = (80 x 1.73^2 / 2) + (65 x 10^2 / 2)

Kf = 3370 J

Ki is not equal to Kf.

hence kinetic energy not conserved.

delta(K) = Kf- ki = 129 J

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