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Period for SHM of a block on a spring shot by a bullet

  • Thread starter ggorrilla
  • Start date
  • #1
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So I've been working on this problem for a while and can't seem to come up with anything and I don't really understand why I'm wrong in what I did. I could just get the answer online, but I actually want to understand what's happening.

Homework Statement



A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0cm. After the impact, the block moves in SHM. Calculate the period of this motion.

Homework Equations



(Period) T = 2pi*sqrt(m/k)
(Kinetic Energy) K = .5*m*v^2
(Potential Energy) U(spring) = .5*k*x^2
(Momentum) P = mv

The Attempt at a Solution


I'll use m for the mass of the bullet and M for the mass of the block.

So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).
Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)
v(final) = 2.182514m/s

So then I assumed that the energy that the system has right after the bullet makes contact is fully kinetic, since in SHM all of the energy is kinetic when at the origin. I think I'm right in assuming this, but maybe that's where I went wrong. But anyways, continuing with what I took as the assumption, I then used the formula for kinetic energy.
K = .5*(M+m)(v(final))^2 So,
K = .5*(.922kg+.00812kg)*(2.182514m/s)^2 = 2.214492J

So then I went and assumed that all energy would be potential when the spring is fully compressed.

U = .5*k*x^2
2.214492J = .5*k*(.18m)^2
k = 136.697066N/m

So then I simply plugged the k into the formula for period and got:
T = 2*pi*sqrt((M+m)/k)
T = 2*pi*sqrt((.922kg+.00812kg)/136.697066N/m)
T = .518286s

Apparently this solution is wrong though, so I was wondering if anyone could help me out so that I could better understand this stuff. : /
 
Last edited:

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46
So I've been working on this problem for a while and can't seem to come up with anything and I don't really understand why I'm wrong in what I did. I could just get the answer online, but I actually want to understand what's happening.

Homework Statement



A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0cm. After the impact, the block moves in SHM. Calculate the period of this motion.

Homework Equations



(Period) T = 2pi*sqrt(m/k)
(Kinetic Energy) K = .5*m*v^2
(Potential Energy) U(spring) = .5*k*x^2
(Momentum) P = mv

The Attempt at a Solution


I'll use m for the mass of the bullet and M for the mass of the block.

So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).
Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)
v(final) = 2.182514m/s

So then I assumed that the energy that the system has right after the bullet makes contact is fully kinetic, since in SHM all of the energy is kinetic when at the origin. I think I'm right in assuming this, but maybe that's where I went wrong. But anyways, continuing with what I took as the assumption, I then used the formula for kinetic energy.
K = .5*(M+m)(v(final))^2 So,
K = .5*(.922kg+.00812kg)*(2.182514m/s)^2 = 2.214492J

So then I went and assumed that all energy would be potential when the spring is fully compressed.

U = .5*k*x^2
2.214492J = .5*k*(.18m)^2
k = 136.697066N/m

So then I simply plugged the k into the formula for period and got:
T = 2*pi*sqrt((M+m)/k)
T = 2*pi*sqrt((.922kg+.00812kg)/136.697066N/m)
T = .518286s

Apparently this solution is wrong though, so I was wondering if anyone could help me out so that I could better understand this stuff. : /
I would certainly go about this problem the same way you have - so I would suggest re-checking your calculations in case there is a slip somewhere.
 
  • #3
ehild
Homework Helper
15,394
1,801

Homework Statement



A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg ...


The Attempt at a Solution


I'll use m for the mass of the bullet and M for the mass of the block.

So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).
Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)
/
Is the mass of the block 0.992 kg or 0.922 kg?

ehild
 
  • #4
2
0
AAHHH!!!!! I can't believe that I didn't catch that, I reworked the entire problem for hours and couldn't find a way to get .557s, which was apparently the answer, but that was my mistake every single time. The correct mass is .992kg, I feel like such an idiot, thank you so much for the help!
 
  • #5
ehild
Homework Helper
15,394
1,801
AAHHH!!!!! I can't believe that I didn't catch that,
Such things happen with everybody...:smile:


ehild
 

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