Question

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is ?k.

Part A

Find an expression for the bullet's speed vbullet.

Part B

What is the speed of a 9.0 g bullet that, when fired into a 12 kg stationary wood block, causes the block to slide 5.4 cm across a wood table? Assume that ?k=0.20.

Answer 1

Part A: The bullet of mass m fired into the block of mass Mand the block moves the distance d is as shown in the figure. :M+m:d The initial momentum of the bullet is mv, and the final momentum of the bullet is zero, because after the firing, the bullet gets stuck into the block. Initially the block is at rest. So, the initial momentum of the block is zero. After the firing the block moves the distance d with the speed v\ So, the final momentum of the block is (m+M)V The law of conservation the momentum is m+M According to the work-energy theorem, the change in the kinetic energy is equal to the work done by the frictional force (f;)

The work done by the frictional force after the bullet strikes the block. So, the initial kinetic energy(KE ) is-(m+M)% The final kinetic energy (KEz) is zero, because the block with bullet comes to rest after 2 travelled the distance d. The frictional force (fk) is negative because, it is opposite to the motion. The work-energy theorem is m+M Substitute 1y2/%gd for Vit in the equation v,- v* and solve for v m+M 2Hgd Therefore the speed of the bullet is, m+M 2Hkgd

Part B Calculate the speed of the bullet as follows: 448 023020/osaps 6.4c) 1( 9.0g+12kg 9.0g ( 9.0 x 10-3 kg+12 kg 9.0 × 10-3 kg (2)(0.20)(9.8 m/s. )(5.4×10-2 m) v 614.0m/s