A 10g bullet strikes and becomes embedded in a 0.200kg block of wood placed on a

Question

A 10g bullet strikes and becomes embedded in a 0.200kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of sliding friction between the block and the surface is 0.4 and the impact drives the block a distance of 8m before it comes to rest, what is the velocity of the bullet before it strikes the block of wood?

I've tried 2 different ways to do it which are:

1.)

m1=0.010kg
m2=0.200kg
s=8m
µ=0.4

1. E(kinetic)=E(frictional)
1/2(m1+m2)v^2=µ(m1+m2)gs
1/2(0.210)v^2=0.4(0.210)(9.8)(8)
v2=7.92m/s

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

2.)

ma=µmg

a=µg
a=0.4*9.8
a=3.92

v^2=0
a=-3.92
s=8m
u^2=?

v^2=u^2+2as
0=u^2+2(-3.92)(8)
u^2=7.92

m1v1=(m1+m2)v2
0.010v1=0.210*7.92
v1=166.32m/s

I'd love to know if I'm on the right track here

Explanation / Answer

Yeah, both the methods are right

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