Simple harmonic motion: mass on a spring is hit by a bullet

  • #1
7
0

Homework Statement


A 4.0kg block is suspended from a spring with force constant of 500N/m.
A 50g bullet is fired into the block from directly below with a speed of 150m/s and is imbedded in the block.
Find the amplitude of the resulting simple harmonic motion.

Homework Equations



F=-kx
E_p=mgh
E_k=(mv^2)/2
E_pspring=(kx^2)/2
m1v1=m2v2

The Attempt at a Solution



Using conservation of momentum:
0.05(150)=(4+0.05)v2
v2=1.852 m/s

MLYBAGY.png

I am lost on how the decide where the equilibrium position is. Is it the resting position with the block considered the equilibrium position, or is it without the block, or is it with the block and the bullet?

I said that the kinetic energy
0.5*(4.05)(1.852)^2 is equal to 0.5k(x)^2 and completely ignored mgh(potential energy). Then i assumed x is the amplitude. This seems incorrect of course.

Any help would be appreciated, thanks!
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,594
5,991
The new equilibrium position will, by definition, be wherever the block with embedded bullet would hang at rest. How does that change your answer?
 
  • #3
7
0
The new equilibrium position will, by definition, be wherever the block with embedded bullet would hang at rest. How does that change your answer?
I'm under the assumption that at equlibium
The new equilibrium position will, by definition, be wherever the block with embedded bullet would hang at rest. How does that change your answer?
I am confused on whether i should consider there is already a spring potential on the block, before hit by the bullet, or if it has no spring potential at that point
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,594
5,991
I am confused on whether i should consider there is already a spring potential on the block, before hit by the bullet, or if it has no spring potential at that point
The spring is not relaxed, so it certainly has potential energy then.
 
  • #5
7
0
I used the position at block+spring as the equilibrium point, and solved it using two methods to get an answer. Even though i didnt consider the block+spring potential energy, since that was my equilibrium point, would my answer be correct?

0G6jxxP.jpg

BnmbBHJ.jpg


Sorry about the pictures instead of latex. I will learn it someday :D
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,594
5,991
I used the position at block+spring as the equilibrium point, and solved it using two methods to get an answer. Even though i didnt consider the block+spring potential energy, since that was my equilibrium point, would my answer be correct?

0G6jxxP.jpg

BnmbBHJ.jpg


Sorry about the pictures instead of latex. I will learn it someday :D
I would take into account that the spring length on impact of bullet is different from the equilibrium length for block+bullet. You can determine the spring PE and GPE of each of those two positions. That will give you a little more KE when subsequently at the equilibrium length. (You can see that by considering what would happen if the bullet had no speed when it stuck to the block. The block+bullet would bob up and down a little.)
 
  • #7
7
0
I think i see what you mean. I will try to include the GPE as well and see what i end up with.
Thanks!
 

Related Threads on Simple harmonic motion: mass on a spring is hit by a bullet

Replies
3
Views
8K
Replies
2
Views
1K
Replies
10
Views
5K
Replies
3
Views
5K
Replies
7
Views
1K
Replies
7
Views
92K
Replies
1
Views
892
Replies
2
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
6
Views
10K
Top