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Old Apr 15th 2009, 12:13 PM   #1
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Bullet impact

A 12 g bullet is fired into a 100 g block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.65, what is the speed of the bullet immediately before impact?
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Old Apr 16th 2009, 10:07 AM   #2
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M = mass of the block
m = mass of the bullet
u = velocity of the bullet before impact
U = velocity of block+bullet just after the impact.
a = accelleration of the block+bullet after impact
k = coefficient of kinetic friction
g = accln due to gravity.
x = distance travelled by block+bullet (=7.5 m) before comming to rest.

The friction force = - k(M+m)g
hence acceln of block+bullet = -kg


$\displaystyle
V^2 = U^2 + 2ax
$
$\displaystyle 0 = U^2 -2kgx$

U = sqrt{2kgx}

Now we use law of conservation of momentum

mu = (M+m)U =(M+m)*sqrt{2kgx}
u = (M+m)*sqrt{2kgx}/m
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