# A 66-kg skier, initially at rest, slides down a hill 25 m high, then has a completely inelastic collision with a stationary 72-kg skier. Fricition is negligible. What is the speed of the skiers immediately after the collision? The 66 kg skier slides down a hill 25 m high. The potential energy of the skier at the top of the hill that has a height of 25 m is given by PE = m*g*h = 66*9.8*25. At the bottom of the hill all the potential energy is converted to kinetic energy and is equal to KE = (1/2)*m*v^2 = 33*v^2

Equating the two gives: 33*v^2 = 16170

=> v = sqrt 490

The momentum of the 66 kg skier is 66*sqrt 490. The initial momentum of the 72 kg skier when the 66 kg skier collides is 0. The total momentum is conserved during the collision. Let the speed of the skiers after the collision be equal to V.

66*sqrt 490 + 0 = (66 + 72)*V

=> V = 66*sqrt 490/138

=> V = 10.58 m/s

The speed of the skiers after their completely inelastic collision is 10.58 m/s.

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