# Oscilations: Bullet collides with wood block on spring

1. Dec 3, 2013

### nickb145

1. The problem statement, all variables and given/known data

A 6.20g bullet traveling at 485m/s embeds itself in a 1.73kg wooden block at rest on a frictionless surface. The block is attached to a spring with k = 79.0N/m

A)Find the Period
B)Find the Amplitude
C)Find the totla energy of the bullet+block+spring system before the bullet enters the block.
D)Find the total energy of the bullet+block+spring system after the bullet enters the block.

2. Relevant equations
T=2∏$\sqrt{}k/m$ or T=2∏ω
K=1/2mv
P=1/2kx
X=Acos(ωt)
Vx,max=ωA

3. The attempt at a solution

Ok so i found the Time T=2πsqrt(1.7362/79)== .931s

Now i'm literally stumped with finding the amplitude. I've tried Vmax=ωA and a few other ways and can't understand what i'm doing incorrectly.

Last edited: Dec 4, 2013
2. Dec 3, 2013

### voko

What's wrong with $V_{max} = \omega A$?

3. Dec 3, 2013

### nickb145

I'm not really sure. I used it because i assume that 485 IS the max velocity

4. Dec 3, 2013

### voko

That was the velocity before impact. What the velocity of the block with the embedded bullet immediately after impact?

5. Dec 3, 2013

### nickb145

I think using the conservation of momentum will work.

M1V1 + M2V2 = (M1 + M2)V3
.0062(485)+1.73(0)=(1.7362)V3 V=1.73m/s (that is a massive drop)

6. Dec 3, 2013

Yep.

7. Dec 3, 2013

### nickb145

Interesting.
SO i just take the difference between Vmax and Vmin?
Vmax 485=Sqrt(79/.0062)A A=4.296
Vmin 1.731=-sqrt(79/1.7362)A= -.256

Which i now have is 4.039m. Which mastering physics tells me is incorrect.

8. Dec 3, 2013

### voko

The velocity the bullet had before impact has no significance for oscillations. Important is only the velocity of the block. What is the max velocity of the block after impact?

9. Dec 3, 2013

### nickb145

1.73m/s?

10. Dec 3, 2013

### voko

Yes.

11. Dec 4, 2013

### nickb145

So now the Amplitude is .256m using Vmax=sqrt(k/m)A

12. Dec 4, 2013

### voko

Looks good.

13. Dec 4, 2013

### nickb145

Now for the energy before the bullet strikes.
Before the bullet enters the wood block i just have the kinetic energy of the bullet. Because i assume the spring is at x=0 and that just cancels out the potential energy.

Now for after the collision I used the X=Acos(ωt) to get X. I got .2544 by adding the two masses, used the time and amplitude i originally had.

1/2(1.7362)(1.73)^2+1/2(79)(.2544) I tried this but getting something incorrect. Is it because of what i did to the kinetic energy?

14. Dec 4, 2013

### nickb145

Never mind i figured it out. Thank you very much for the assistance!