# Bullet fired from below into block suspended by spring

• Swedishfish
In summary, a 4kg block is suspended from a spring with a 500N/m spring constant. A .05kg bullet is fired from below with a speed of 150m/s and comes to rest in the block. The amplitude of the resulting simple harmonic motion is found to be 113.9J.
Swedishfish

## Homework Statement

A 4 kg block is suspended from a spring with k=500N/m. A .05kg bullet is fired from below into with a speed of 150m/s and comes to rest in the block.
a.)Find the Amplitude of the resulting simple harmonic motion.
b.)What fraction of the original KE of the bullet appears as mechanical energy in the harmonic oscillator?
^^^This is the problem word for word^^^

KE=1/2mv^2
F=-ky
U=1/2ky^2
T=2pi(m/k)^1/2
P cons.
E cons.
etc...

## The Attempt at a Solution

Where I am atm is, the problem basically has two parts, the collision which is completely inelastic so Pcons applies and then the compression of the spring and oscillation of the combined masses.

setting up the equations for the collision I apply Pcons.
MblkVblk+MbtVbt=MblkVblk+MbltVblt --> Vbb=(MblkVblk+MbtVbt)/(Mblk+Mblt)
-->Vbb=1.85m/s

after the collision having Mbb=4.05 and Vbb=1.85m/s the combined block and bullet mass has
KEbb=113.9J

now I know that at the point of maximum compression of the spring, the mass will be at Vbb=0. So, Applying E cons I set KEbb=Us

113.9J=1/2ky^2 -->227.8=ky^2 --> 227.8/k=y^2 --> (227.8/k)^1/2=y...

Now this is where confusion sets in for me. If I take the problem as I have been, and continue to ignore the hell out of gravity, then there is nothing to dampen the oscillations over time, and I've basically got amplitude since I just solved for maximum compression of the spring. However if I let gravity in the door, things get more complex. This problem is part of a huge packet of problems I'm doing to prep for senior exams so it covers everything from intro on up and this particular problem is located smack in the middle of the medium-difficult ranked intro problems so I don't think the intention here is from me to go charging off with a diff-eq into classical mechanics territory. Since they gave me just masses, velocities and the spring constant I'm kinda thinking this is supposed to be a fairly straight forward energy/momentum problem...

If anyone would look this over and see if there is some thing I'm missing here I'd appreciate it very much as it's been almost 4 years since I took intro. Thanks.

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No, don't use gravity. Gravity is already implicitly playing a role. The string has already stretched to a certain spot because of gravity. The only thing that this bullet does is give an initial velocity to the simple harmonic spring oscillator.

$$x(t) = c_1 sin(\sqrt{k/m}t) + c_2 cos(\sqrt{k/m}t$$

Here is a more detailed explanation to leave out gravity (I copied it out of a paper I wrote a while back):
The reason behind gravity not explicitly appearing in equation (1) has to due with building the equation around the idea of the equilibrium point. When the spring and mass are at rest the sum of the forces must all equal zero, and since there is no velocity or acceleration at this point then ks = mg where s is the distance from the ceiling to the mass. The only time a restoring force will be applied is when the spring has been stretched away from this equilibrium point, but if the spring is lifted up from the equilibrium point the force of gravity on the mass will be equal to the restoring force that would occur if it were stretched, precisely because ks = mg. In other words, whether the spring is lifted up or stretched down the forces will be equal; thus, the equation does not need to explicitly use a term for the force of gravity because it is implicit in the restoring force.

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Thanks I figured it was something along those lines but wanted to run it by someone just because it's been so long...every time I do these problems it's like I know how to do it, but the right way is just out of reach.

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## 1. How does the bullet fired from below affect the suspended block?

When the bullet is fired from below, it creates an upward force on the block, causing it to move upwards and compress the spring. This is due to the principle of action and reaction, where the force of the bullet on the block is equal and opposite to the force of the block on the bullet.

## 2. What happens to the energy of the bullet when it hits the suspended block?

When the bullet hits the suspended block, its kinetic energy is transferred to the block and the spring. The block will then move upwards with a higher velocity, while the spring will compress and store potential energy. Some of the bullet's energy may also be lost to other factors such as friction and heat.

## 3. Can the spring break from the impact of the bullet?

It is possible for the spring to break from the impact of the bullet, depending on the strength and material of the spring. If the spring is too weak to withstand the force of the bullet, it may snap or deform. It is important to use a strong and durable spring for this experiment.

## 4. How does the mass of the bullet affect the movement of the suspended block?

The mass of the bullet will determine the amount of force it exerts on the suspended block. A heavier bullet will have a greater force and therefore cause the block to move upwards with a higher velocity. However, the mass of the bullet does not affect the motion of the block after the impact, as the block's movement is determined by the force of the bullet and the strength of the spring.

## 5. What factors can affect the accuracy of this experiment?

There are several factors that can affect the accuracy of this experiment, including the height from which the bullet is fired, the angle at which the bullet hits the block, the strength and material of the spring, and external factors such as wind and air resistance. It is important to control these variables as much as possible to obtain accurate results.

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