Brian was coasting over a level frictionless ice surface riding on a 10 kg sled

Question

Brian was coasting over a level frictionless ice surface riding on a 10 kg sled loaded with himself of mass 50 kg and five identical frozen Turbot which has mass 2 kg each fish. He was being pursued by the Spanish fleet armed with fine mesh nets. His speed was 10 m/s, and he had to get over a 6m high hill to escape. He carried a sling shot and could shoot the Turbot backwards with a horizontal speed of 20 m/s relative to his sled. Assuming he could shoot one fish at a time, and finished all his shots before he reached the hill, calculate the minimum number of Turbot he had to shoot in order to escape over the hill.

Explanation / Answer

In order to get over the hill his velocity should be: mgh = 1/2mv^2 (9.8)(6m) = (0.5)v^2 v = 10.844m/s If he is carrying 5 turbots, his initial mass is: 10kg + 50kg + 5(2kg) = 70kg So, when he throws one, by conservation of momentum: (70kg)(10m/s) = (68kg)v + (2kg)(-10m/s) because the 20m/s backwards is relative to the sled. v = 10.588m/s So he throws another: (68kg)(10.588m/s) = (66kg)v + (2kg)(-9.412m/s) v = 11.194m/s He has to throw two.

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