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Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. On one particular shot, his 1.2 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is 64 m/s and the cheese has a mass of 123 g, what is the coefficient of friction between the cheese and ice?
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Added Mon, 03 Aug '15
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given :
muzzle velocity = 64 m/sec

so.. velocvity of bullet just before hitting the cheese = 64 m/sec

momentum of bullet just before hitting chese = mass of bullet * velocity = 1.2 * 64 = 76.8 g * m/sec

after sticking in cheese... let the velocity of the bullet + cheese be v ...

so.. momentum after collision = (123+1.2)*v

now.. conserving momentum :
124.2 * v = 76.8

so.. velocity of cheese + bullet = 76.8 / 124.2 = 0.61836 m/sec ..

so.. kinetic energy of cheese + bullet system = 0.5 * mass * v^2 = 0.5 * 0.1242*0.61836^2 = 0.023745 J


let the coefficient of friction be mu ....

now.. friction force = mu * mass *g = mu * 0.1242 * 9.81

work done by friciton = friceion force * distance = 0.1242 * 9.81 * 0.25 * mu


now...
work done by friction = kinetic energy of bullet + cheese

so... 0.1242 * 9.81 * 0.25 * mu = 0.023745

so. coefficeint = mu = 0.077954
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Added Mon, 03 Aug '15
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