Calculate the mass of the block of wood

[davewisniewsk]

Homework Statement

A 34.0 g bullet traveling at 122 m/s emdeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring, k=99.0 N/m, a maximum of 1.20 cm. Calculate the mass of the block of wood.

Homework Equations

Eelasticpot= 1/2 k x^2
Ekinetic = 1/2 m v^2
m1v1=(m1+m2)(v2)

The Attempt at a Solution

I've looked into this indepth online and found similar questions, but nothing where mass is the unknown. I have used law of conservation of energy

ie. 1/2(0.034)(122)^2 = 1/2(0.034 + m)v^2 = 1/2(99)(0.012) to find the velocity of the block with the bullet lodged in it, (and likewise using my knowns to find the mass of the block), however this results in a rediculous 1207 kg, which seemed far too high.

So, I need to know a better way to do this. One thing to point out (perhaps a flaw in my attempts), is that the kinetic energy of the bullet (before becoming enlodged) does not equal the elastic potential energy of the spring. Could this mean it is not a closed system.

Thanks for any help, I know I'm new here but my friend and I have been working on this problem all weekend with no luck.

 

[alphysicist]

Hi davewisniewsk,

Homework Statement

A 34.0 g bullet traveling at 122 m/s emdeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring, k=99.0 N/m, a maximum of 1.20 cm. Calculate the mass of the block of wood.

Homework Equations

Eelasticpot= 1/2 k x^2
Ekinetic = 1/2 m v^2
m1v1=(m1+m2)(v2)

The Attempt at a Solution

I've looked into this indepth online and found similar questions, but nothing where mass is the unknown. I have used law of conservation of energy

ie. 1/2(0.034)(122)^2 = 1/2(0.034 + m)v^2 = 1/2(99)(0.012)

This equation is assuming that mechanical energy is conserved for the block/bullet/spring system for the entire motion, which is not true. Instead, think about the two different parts of the motion: the collision, and then the sliding. What is conserved during the collision? What is conserved during the sliding? The answers to those questions will tell you how to set up equations for each part.

 

[davewisniewsk]

Thanks!

I'm guessing that what you hinted at was that in case one- bullet hits block, momentum is conserved.

I rearranged this and solved for the velocity of the block/bullet pair in terms of the bullet's mass, the blocks unknown mass and the initial bullet velocity.

Then, I used law of conseravation of energy to look at the block hitting the spring.

Again, I solved for velocity of the block/bullet, but this time in terms of spring constant, compression and the masses.

I subbed them together and solved.

This is where my problem is: my answer (which I have got on numerous occasions) seems far to large to be true. I come up with approx. 1208 kg, for the mass of the block of wood.

Could this be possible and maybe the question is supposed to yield such a large answer?

 

[alphysicist]

Right, I was just commenting on the equation in your post, not your answer. Your answer looks right to me (I get a slightly different answer probably due to rounding during the calculation).