# Simple Harmonic Motion

## Homework Statement

A 14.3 g bullet is fired into and embeds itself
in a 2.76 kg block attached to a spring with a
spring constant of 18.4 N/m and whose mass
is negligible.
How far is the spring compressed if the
bullet has a speed of 247 m/s just before it
strikes the block, and the block slides on a
frictionless surface?

## Homework Equations

.5(m)(v^2) + .5 (k)(x^2) = .5(k)(A^2)
velocity = (omega)(amplitude)
momentum = m1v1

## The Attempt at a Solution

Here's my process but its not right because I got square root of a negative number:
I took the v=wA and solved for Amplitude. But to get the velocity of the wood, I did momentum of bullet = momentum of block. I plugged in my velocity(v2) into the formula and solved for amplitude. I then used the .5(m)(v^2) + .5 (k)(x^2) = .5(k)(A^2) and solved for x. To find the velocity, i assumed the velocity was velocity of block + bullet, in the equation, I set up another momentum equation: momentum of bullet + momentum of block = momentum of bullet + block and solved for velocity of block+ bullet. I then used this new velocity to find the distance, x. The only problem is that I got a square root of a negative number, which is obviously wrong. What am i doing wrong?

## The Attempt at a Solution

hello yang, I don't know why you have posted all of that after your question.

Also, I have concerns about your question, it does not seem to be about simple harmonic motion, it seems to be about the conversion of momentum, and the conservation of energy.

to find the A value you are looking for, you do not start with w = vA. You start off the with energy in this bullet. This energy should be the maximum energy that is eventually stored inside the spring. Once you have the energy inside the spring, you can figure out how much it is compressed. Then once after you have done that, you will be able to answer a possible follow up question about the harmonic motion.

The circumstances of this problem could involve harmonic motion, but you have asked no questions with regard to that.

So if you're saying I have to find the maximum energy in the bullet, would I find it using the Kinetic energy formula? Since the energy in the bullet has kinetic energy, would I use .5mv^2 to find the energy in the bullet? And with that, how would I find the energy of the spring?

yes thats right, you use .5mv². The formula for energy inside a spring with a spring constant of k and a distance compressed or stretched x is 0.5kx². so you would essentially be setting those two equations equal to each other and finding x.

But how would I incorporate the block into the equation since the energy is first transferred from the bullet to the block and then to the spring.

hmm good question. I don't think it matters.

From a momentum standpoint, and an energy standpoint, the block and the bullet have the same values for p and E as the bullet did before it hit the block. this suggests that there is no difference in the amplitude of the spring's oscillations.

But if the energy is being transferred from the bullet to the block, then the total amount of energy given to the spring would be the energy from the block + bullet right?

If i did it your way, I assume that the energy given to the spring is the energy from the bulllet. So to calculate the energy of the bullet, I used .5mv^2, using the mass of the bullet, .0143 kg and velocity of the bullet, 247. I plugged it into the equation and solved for x but got the wrong answer. Am I calculating wrong or something?

what is your answer? I got 6.9m is what it looks like. which seems like a lot, but the spring constant is very low, so that seems plausible.

same here. I got 6.885826258 which rounds to 6.9, but its wrong. So since you got the same answer, then I'm assuming that the energy from the block is needed then. We just cant assume that the energy from the bullet = energy of block

Have you both tried finding the velocity of the block after the bullet is embedded?
(B)is for block and (b) is for bullet:

$$m_{b}v_{b} + m_{B}v_{B} = m_{Bb}v_{Bb}$$

$$v_{Bb} = \frac{m_{b}v_{b} + m_{B}v_{B}}{m_{Bb}}$$

I get an answer for this around $$1.273ms^{-1}$$

...

I was going somewhere with this, but I forgot =D

Last edited:
Thats the way I originally did it, but you first had to find the velocity of the block before collision. That was the part that messed me up because I didn't know how to solve for the velocity of the block before collision. how did you find the velocity of the block before collision?

I assumed the velocity of the block before the collision was zero.

edit:

Do you know what the answer should be?

Using your way Cilabitaon, I gota velocity of bullet+block to be 1.373149984 because you're just multiplying .0143kg with 247m/s and dividing that answer with the mass of the block+bullet, 2.7743kg. With the velocity of the block + bullet, I plugged it into this formula: .5mv^2 = .5kx^2 and solved for x. I got x = .494364604. This what you got?

Using your way Cilabitaon, I gota velocity of bullet+block to be 1.373149984 because you're just multiplying .0143kg with 247m/s and dividing that answer with the mass of the block+bullet, 2.7743kg. With the velocity of the block + bullet, I plugged it into this formula: .5mv^2 = .5kx^2 and solved for x. I got x = .494364604. This what you got?

Exactly; except you misquoted your velocity I think. =]

Cilabitaon, thanks for the help, but can you help me with one more question. I have another post: Need help with Simple Harmonic Motion problem and I cant seem to get part 4. thanks

I want to talk about the fundementals of this. No energy is lost anywhere in this system, there is no friction. Unless you claim that there is energy lost to heat as the bullet penetrates the wood block. The question says nothing of this

Therefore we have to assume that all the energy from the bullet is conserved and put into the spring. The mass of the block will take away a lot of velocity, but it will have the same momentum as the bullet had. Unless its an elastic collision which i doubt.

Is there any argument mounted that there is less energy in the spring system at the end, than there was in the bullet at the beginning?

Is there any argument mounted that there is less energy in the spring system at the end, than there was in the bullet at the beginning?

No, the collision is 100% elastic in this model.

So then 6.9 has to be correct

So then 6.9 has to be correct

Would you mind showing your working?

no problem.

0.5mv² = 0.5kx²

x = sqrt(mv²/k)