A 36-kg child is moving on a 2.0-kg skateboard at speed 6.0 m/s when she comes t

Question

A 36-kg child is moving on a 2.0-kg skateboard at speed 6.0 m/s when she comes to a ledge that is 1.2 m above the surface below. Just before reaching the ledge, she pushes off the board. The board leaves the ledge moving horizontally and lands 8.0 m horizontally from the edge of the ledge.

A. Determine the launch speed of the skateboard.

B. Determine the initial momentum of the child-skateboard system.

C. Determine the speed of the child after leaving the ledge.

D. Determine the kinetic energy of the child after leaving the ledge.

Explanation / Answer

part A )

time to fall t = sqrt(2h / g)

t = sqrt(2 * 1.2m / 9.8m/s²) = 0.495 s

Therefore launch speed v = d / t = 8m / 0.495s = 16.17 m/s = 16 m/s

Part B )

initial momentum = P = (Mc + Msb) * vi

P = ( 36 + 2 ) * 6 = 228 kg . m/s

part c )

Now conserve momentum

initial p = final p

(36 + 2.0)kg * 6.0m/s = 36kg * v + 2.0kg * 16m/s

v = 196 / 36 = 5.4 m/s

part D )

KE = 1/2*mv^2 = 1/2 * 36kg * (5.4m/s)² = 530 J

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