Speed Of A Bullet Before It Strikes A Block Of Wood

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SUMMARY

The discussion centers on calculating the velocity of a bullet before it strikes a block of wood using principles of physics. The bullet has a mass of 10g and strikes a 0.200kg block, with a coefficient of sliding friction of 0.4. Two methods were employed to arrive at the bullet's velocity, both yielding a result of 166.32 m/s. The calculations utilized the equations for kinetic energy and frictional force, confirming the accuracy of the methods used.

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amber87
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Homework Statement



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?



Homework Equations



E(kinetic)=E(frictional)
1/2(m1+m2)v2^2=μ(m1+m2)gs

or

ma=μmg

a=μg

v^2=u^2+2as

v1(m1)=v2(m1+m2)


The Attempt at a Solution



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. Thanks a million in advance.
 
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amber87 said:

Homework Statement



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?



Homework Equations



E(kinetic)=E(frictional)
1/2(m1+m2)v2^2=μ(m1+m2)gs

or

ma=μmg

a=μg

v^2=u^2+2as

v1(m1)=v2(m1+m2)


The Attempt at a Solution



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. Thanks a million in advance.

I would have used your method 1. and agree with your calculations.
There is a chance, that if this was an introductory physics question, that you were only required to use g = 10. This makes calculations much simpler and able to be done without a calculator. In that case that first velocity is 8, and the initial velocity this 168 m/s,
 

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