A 35.0 gram bullet strikes and embeds itself in a 1.90 kg block that is initiall

Question

A 35.0 gram bullet strikes and embeds itself in a 1.90 kg block that is initially at rest. The block slides 4.00 m across the floor before coming to a stop.

a) The coefficient of kinetic friction between the block and the floor is mk=0.4. What is the work done on the block by kinetic friction as the block slides across the floor?

b) What was the initial velocity of the block when it began sliding across the floor (that is immediately after the bullet embeds in the block)?

c) What was the speed of the bullet immediately before it struck the block?

d) Suppose that instead of sliding across a frictionless floor, the block was allowed to slide up a frictionless incline after the bullet is embedded. To what maximum height does the block slide up the incline?

Explanation / Answer

a) work done =-F.s

=-(1.9+0.035)*9.81*4

=-75.93 J

b)let the velocity of the system after collision be v.so,

0.5*(1.9+0.035)*v^2 = 75.93

or v=8.85 m/s

so, conserving momentum,

0.035*u=(1.9+0.035)*8.85

or u=489.279 m/s

c)conserving energy,

0.5*(1.9+0.035)*8.85^2 = (1.9+0.035)*9.81*h

or h=4 m

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