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 μk​

A) What is the speed of a 12 g bullet that, when fired into a 9.0 kg stationary wood block, causes the block to slide 5.6 cm across a wood table? Assume that μk=0.20

Express your answer to two significant figures and include the appropriate units.

Answer 1

let Pi= momentum of bullet Pb

Pf= momentum of bullet+block

P=0, since there are no external forces

Pi=Pf

mv=(M+m)V

v=(M+m)V/m...............eqn1

we know the formula from newtons law

Fnet=m*a

from the free body diagram

Fnet_x= -f=ma_x

f=N

so, -N=ma_x

Fnet_x=N-W=ma_x=0

N=W=mg

a_x=-f/m

= -k mg/m

=-k *g

=-0.2*9.8= -1.96 m/s²

using the kinematics equation

v²=V²+2ax

V=v²-2ax

=0-2*(-1.96m/s²)*(0.056m)

= - 0.47m/s

putting this value in eqn 1

v=(M+m)V/m

=(9kg+0.012kg)*0.47m/s/0.012kg

=352.97 m/s

so the speed of the bullet=352.97m/s