Calculating Initial Speed of Bullet After Collision with Spring-Mounted Block

[PiRsq]

Friction is negligible

A bullet traveling at speed v hits a wooden block which is attached to a spring with constant 200 N/m. After the impact if the spring compresses to a maximum of 0.1m and the mass of the bullet is 0.01kg and the mass of the block is 0.5kg find the initial speed of the bullet?

This was on my test today and I didn't get the answer, how would you go about doing this question?

 

[Doc Al]

Why not show us how you tried to solve it?

Hint: you'll need to use both conservation of momentum and conservation of energy.

 

[PiRsq]

I found the Elastic potential energy first to see the energy of the whole system by using:

1/2kx^2


Then since the kinetic energy must equal the total energy before the spring is stretch, I did:

1/2kx^2=1/2mv^2 and I solved for V

 

[Doc Al]

Originally posted by PiRsq
Then since the kinetic energy must equal the total energy before the spring is stretch, I did:

1/2kx^2=1/2mv^2 and I solved for V

That's the speed of the block plus bullet---after the collision. Now find the speed of the bullet before the collision. (Re-read my previous hint.)

 

[PiRsq]

So momentum is conserved...


mv1+mv2=mv1'+mv2'
v1'=v2'=v'
v2=0
mv1=v'(m1+m2)


So I think to get v':

1/2kx^2=1/2mv^2

The energy came out to 1 Joule.

Equation I joule with kinetic energy :

2=(m1+m2)v^2
v=sqrt(2/m1+m2)

Now that I think equals v' right?

Subbing that back into the equation we can solve for v1

Is that somewhat right?

 

[Doc Al]

It is exactly right. Good job!

 

[PiRsq]

Thanks for the help Doc I just wish I thought of that during the test