Oscilations: Bullet collides with wood block on spring

  • Thread starter nickb145
  • Start date
  • #1
68
0

Homework Statement



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.

Homework Equations


T=2∏[itex]\sqrt{}k/m[/itex] or T=2∏ω
K=1/2mv
P=1/2kx
X=Acos(ωt)
Vx,max=ωA


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:

Answers and Replies

  • #2
6,054
391
What's wrong with ##V_{max} = \omega A##?
 
  • #3
68
0
What's wrong with ##V_{max} = \omega A##?

I'm not really sure. I used it because i assume that 485 IS the max velocity
 
  • #4
6,054
391
That was the velocity before impact. What the velocity of the block with the embedded bullet immediately after impact?
 
  • #5
68
0
That was the velocity before impact. What the velocity of the block with the embedded bullet immediately after impact?

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
6,054
391
Yep.
 
  • #7
68
0
Yep.

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
6,054
391
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
68
0
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?

1.73m/s?
 
  • #10
6,054
391
Yes.
 
  • #12
6,054
391
Looks good.
 
  • #13
68
0
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
68
0
Never mind i figured it out. Thank you very much for the assistance!
 

Related Threads on Oscilations: Bullet collides with wood block on spring

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
7K
  • Last Post
Replies
3
Views
10K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
8
Views
5K
Replies
2
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
4
Views
7K
Replies
2
Views
3K
Replies
8
Views
2K
Top