What is their speed at the bottom of the hill if the total vertical drop is 18.5 m?

Question

Gayle runs at a speed of 3.45 m/s and dives on a sled, initially at rest on the top of a frictionless snow covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 18.5 m? Gayle's mass is 50.0 kg, the sled has a mass of 5.00 kg and her brother has a mass of 30.0 kg.
m/s

Answer 1

The first collision of Gayle onto the sled is conservation of momentum for an inelastic collision

3.45*50=v*55
solve for v

v=3.136 m/s

For the vertical drop use energy without friction

55*9.81*5+.5*55*3.136^2=
.5*55*v^2

solve for v

v=10.4 m/s

Now there is another collision with her brother that is inelastic

55*10.4=v*85
solve for v

v=6.73 m/s

They then descend another 13.5 meters together, which is again conservation of energy

85*9.81*13.5+.5*85*6.73^2=
.5*85*v^2

solve for v

v=17.6 m/s

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