Question

# A pistol fires a 0.0050-kg bullet with a muzzle velocity of 1000.0 m/s. The bullet then...

A pistol fires a 0.0050-kg bullet with a muzzle velocity of 1000.0 m/s. The bullet then strikes a 10.0-kg wooden block resting on a horizontal frictionless surface and becomes embedded in the block. The block and bullet then slide across the surface. What was work done on the block and the impulse delivered to the bullet during the collision?

a. 250J, 12.5N-s

b. 12J, 12.5N-s

c. 1.2J and -5.0N-s

d. 0.15J, -2.5N-s

e. 38J, 7.5N-s

as there is no external force, the linear momentum is conserved.

hence momentum before collision=momentum after collision

if speed after the bullet striked the block is v m/s,

then 0.005*1000+10*0=(0.005+10)*v

==v=0.49975 m/s

then impuse delived to the bullet=change in momentum

=mass*(final velocity-initial velocity)

=0.005*(0.49975-1000)=-4.9975 N.s

then work done on the block=gain in kinetic energy of the block

=0.5*mass*(final speed^2-initial speed^2)

=0.5*10*(0.49975^2-0^2)=1.24875 J

hence option c is correct.

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