A bullet of mass 100 grams is fired at a wooden block of mass 2.50 kg that is su

Question

A bullet of mass 100 grams is fired at a wooden block of mass 2.50 kg that is suspended to the ceiling by means of a thin rope of negligible mass and length 1 m. The bullet's initial velocity is 15 m/s, and the block is enough to completely stop this inside. How large is the velocity of the block plus bullet system right after the bullet stopped inside? How high does the block then reach? How fast should a bulHow much work did the block do to stop the bullet? Assuming it took 0.1s for the block to stop the bullet, what was average force exerted by the block on the bullet?

Explanation / Answer

part a:

as there are no external forces, momentum will be conserved.
so initial momentum=final momentum

==>0.1*15+2.5*0=(0.1+2.5)*final speed

==>final speed=0.1*15/(0.1+2.5)
=0.57692 m/s

part b:

height reached by the block=initial speed imparted^2/(2*acceleration due to gravity)

=0.57692^2/(2*9.8)

=0.01698 m

part c:

work done by the block=difference between initial and final kinetic energy

=0.5*0.1*15^2-0.5*(0.1+2.5)*0.57692^2

=10.817 J

part d:

impulse imparted=change in momentum

==>force*time=0.1*(15-0.57692)

==>force=0.1*(15-0.57692)/0.1=14.423 N

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