Conservation of Energy of a bullet

In summary: So, in summary, the conservation of energy equation should include the change in gravitational potential energy, as well as the final kinetic energy of the block and bullet. Ignoring gravity would not give an accurate answer.
  • #1
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hey ppl
A 4.0 kg block is suspended from a spring with a force constant of 500 N/m. A 50g bullet is fired into the block from directly below with a speed of 150 m/s and is imbedded in the block.

So basically... there are 3 sections here.

A. bullet is traveling up and hasn't touched the motionless block yet.

B. bullet impacts the box and is embedded in the box. energy is lost, but momentum is conserved

C. box and bullet travels upward with the resulting momentum from the original bullet.


So.. A to B, it is conservation of momentum. from B to C, it is conservation of energy.

ok... I found the velocity of the block and the bullet when the block is at the equilibrium position. (just when it is impacted) by using the conservation of momentum.

(m1)(vi) = (m1+m2)(vf)

then i use that velocity in the conservation of energy equation. Therefore...

(1/2)(m1+m2)V^2 + 1/2kx^2 = (m1+m2)gh + 1/2KA^2

that may look like a blob to you but my question is simply this.. in this conservation of energy equation... do i include the change in graviational potential energy? I mean.. the block has moved up, right? so it should be included... supposedly.

A couple of my friends disagree with me. (At least according to one of them, he already had a peek at the answer key from the prof and he claims that gravitational potential energy isn't included.)

So I'm confused why it shouldn't be included. Help. Plz. Thx.
 
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  • #2
Feel free to add a term for gravity, but remember that the block will then also be streching the spring a little. IIRC the two will cancel.
 
  • #3
Tell your friends they're wrong.

Well, I guess it depends on the question. If it only asks for the speed at the instant after the collision, use your conservation of momentum equation & you're done.

But beyond that point you are correct in that you must use conservation of energy, & you can't ignore gravity. Take your equation:

(1/2)(m1+m2)V^2 + 1/2kx^2 = (m1+m2)gh + 1/2KA^2
The only thing you left out is the FINAL kinetic energy + (1/2)(m1+m2)Vf^2.
x is the distance the spring has been stretched (you can easily determine this distance from the given information, & substitute the actual value into the equation). Let h = 0 at the position of the block just before the bullet hits, so your A = x - h. (I'm not sure why you capitalized the K on the right side of the equation. I assume that was unintentional, & I'm switching it to lowercase to make it clear that the k on both sides is the same k.

Now your equation becomes:
(1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)k(x-h)^2 + (1/2)(m1+m2)Vf^2
(1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)k(x^2-2xh+h^2) + (1/2)(m1+m2)Vf^2
(1/2)(m1+m2)V^2 + (1/2)kx^2 = (m1+m2)gh + (1/2)kx^2 - xh + (1/2)kh^2 + (1/2)(m1+m2)Vf^2
(1/2)(m1+m2)V^2 = (m1+m2)gh - xh + (1/2)kh^2 + (1/2)(m1+m2)Vf^2
 
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What is the conservation of energy principle?

The conservation of energy principle states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another.

How does the conservation of energy apply to a bullet?

When a bullet is fired from a gun, it has a certain amount of kinetic energy. As it travels through the air, the bullet experiences air resistance and loses some of its kinetic energy. However, this energy is not lost, but is transferred to the air molecules and sound waves around the bullet. When the bullet hits its target, all of its kinetic energy is transferred to the target, causing damage.

What factors affect the conservation of energy of a bullet?

The conservation of energy of a bullet is affected by factors such as the initial velocity of the bullet, the weight and composition of the bullet, and the medium through which it travels (e.g. air, water, or a solid object).

Why is the conservation of energy important in firearms and ballistics?

The conservation of energy is important in firearms and ballistics because it helps us understand the behavior of bullets and how they interact with different materials. This knowledge is crucial for improving the accuracy and effectiveness of firearms, as well as for understanding the physics behind gunshot wounds.

How does the conservation of energy impact the trajectory of a bullet?

The conservation of energy plays a significant role in the trajectory of a bullet. As a bullet travels through the air, it loses energy due to air resistance and gravity. This causes the bullet to slow down and drop in altitude. Understanding the conservation of energy can help ballistics experts calculate the trajectory of a bullet and make accurate predictions about its flight path.

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