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 ve identical frozen Turbot which has mass 2 kg each
sh. He was being pursued by the Spanish
eet armed with ne 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 sh at a time, and nished 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 he must have a velocity of: mgh = 1/2mv^2 (9.8)(6m) = 1/2v^2 v = 10.844m/s So, he needs to increase his velocity by 0.844m/s by throwing "turbots" to escape. I think you said he was carrying five of these things, but I'm not sure. If that is the case then his total mass initially would be: 10kg + 50kg + 5(2kg) = 70kg The first turbot he throws will (by conservation of momentum): 70kg(10m/s) = (68kg)v + (2kg)(-20m/s) v= 10.882m/s increase his velocity to 10.882m/s which should be enough to get over the hill. Of course, if he were carrying more of the "turbots" (I assumed 5 based on the fact that you said "and ve identical frozen Turbot which has mass 2 kg each") then it may take more. Basically, you could repeat this calculation with the additional mass. Just remember, each time he throws one his mass decreases by 2kg.

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