Ballistic Pendulum kinetic energy

In summary: No matter what happens, momentum will ALWAYS be conserved.In summary, a rifle bullet with a mass of 11.0 g and a speed of 360 m/s is fired into a ballistic pendulum with a mass of 9.00 kg and a cord length of 70.0 cm. The initial kinetic energy of the bullet is calculated to be 713 J. The kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded is found to be 1426 J, and the vertical height through which the pendulum rises is determined to be 8.1 m. The conservation of momentum is used to calculate the velocity of the combined bullet and pendulum to be 0.439 m/s. In cases
  • #1
QuarkCharmer
1,051
3

Homework Statement


A 11.0 g rifle bullet is fired with a speed of 360 m/s into a ballistic pendulum with mass 9.00 kg, suspended from a cord 70.0 cm long.

A.) Compute the initial kinetic energy of the bullet

B.) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.

C.) Compute the vertical height through which the pendulum rises.


Homework Equations



The Attempt at a Solution



A.) I just applied 1/2mv^2 to get the initial KE of the bullet to be 713 J (it's correct)

B.) Since the kinetic and potential energy of the pendulum right when the bullet fires is zero, I claim that:

[tex]713 = \frac{1m_{b}v^{2}_{b}}{2} + \frac{1m_{p}v^{2}_{p}}{2}[/tex]
[tex]713 = \frac{(.011)v^{2}_{b}}{2} + \frac{(9)v^{2}_{p}}{2}[/tex]
Now I am assuming that the velocity of the bullet and pendulum are the same since they are now attached so:
[tex]713 = \frac{1}{2}(.011+9)v^{2}[/tex]
[tex]1426 = (.011+9)v^{2}[/tex]
[tex]1426 = (9.011)v^{2}[/tex]
[tex]\frac{1426}{9.011} = v^{2}[/tex]
[tex]\sqrt{\frac{1426}{9.011}} = v[/tex]
[tex]v = 12.6 m/s[/tex]

??

C.)
[tex]\frac{1}{2}(9.011)(12.6)^{2} = (9.011)(9.8)h[/tex]
[tex]h = 8.1m[/tex]

There is no way this is right, the shaft of the pendulum is only 70cm long!
 
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  • #2
For part B, KE is not conserved. So you need to use another law of conservation.

And with v, you know how to get KE.
 
  • #3
Would I use the conservation of momentum?

[tex](0.011)(360) = (9.011)v[/tex]
[tex]v = 0.439 m/s[/tex](Which naturally changed my solution to C to 0.983 cm or 0.00983 m)
 
  • #4
QuarkCharmer said:
Would I use the conservation of momentum?

[tex](0.011)(360) = (9.011)v[/tex]
[tex]v = 0.439 m/s[/tex](Which naturally changed my solution to C)

Yes that is what I got and note B is not asking for v.
 
  • #5
Yeah, now I just pop that velocity (the bullet with pendulum) into the equation of kinetic energy right?

[tex]KE = \frac{1}{2}mv^{2}[/tex]
 
  • #6
QuarkCharmer said:
Yeah, now I just pop that velocity (the bullet with pendulum) into the equation of kinetic energy right?

[tex]KE = \frac{1}{2}mv^{2}[/tex]

Yup. And for C you did as you did with the new KE as you said.

The KE isn't conserved because the bullet used some energy to push itself into the box and in reality some went to friction (heat).
 
  • #7
frozenguy said:
Yup. And for C you did as you did with the new KE as you said.

The KE isn't conserved because the bullet used some energy to push itself into the box and in reality some went to friction (heat).

You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?

Thank you for your help.
 
  • #8
QuarkCharmer said:
You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?

Thank you for your help.

Well I think the bullet would actually have to stop and fall and the box moves forward completely on its own for KE to be conserved..
Because I think also KE isn't conserved because its inelastic and the bullet travels with the box... I'm a little shaky on this so you should confirm it with someone who knows for 100% sure.

But I'm rather sure the bullet would have to stop and not travel with the box.
 
  • #9
QuarkCharmer said:
You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?

Thank you for your help.

Energy is always conserved, but not always the way we want it to be :smile:

Kinetic energy has a tendency to sneak off into all sorts of different avenues like heat, sound, radiation, even potential energy (your bullet could have been 'caught' and its energy used to wind a spring, for example, or by your magnet so that the bullet would then have potential energy in the magnet's field). The energy isn't "gone", it's just not where we want it to be!

Conservation of momentum, on the other hand, can be relied upon.
 

What is a ballistic pendulum?

A ballistic pendulum is a device used to measure the velocity of a projectile by measuring the height to which it raises a pendulum after impact.

How does a ballistic pendulum work?

A projectile is fired into a pendulum, causing it to swing upward. The height to which the pendulum rises is used to calculate the initial velocity of the projectile using the principle of conservation of energy.

What is the principle of conservation of energy in a ballistic pendulum?

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In a ballistic pendulum, the initial kinetic energy of the projectile is transferred to the potential energy of the pendulum at the highest point of its swing.

What factors affect the accuracy of a ballistic pendulum?

The accuracy of a ballistic pendulum is affected by factors such as the mass and velocity of the projectile, the angle of impact, and friction or air resistance. Small errors in measurement and calculation can also affect the accuracy of the results.

What are the practical applications of a ballistic pendulum?

Ballistic pendulums are commonly used in physics experiments and demonstrations to measure the velocity of projectiles. They also have applications in forensic science, where they can be used to determine the speed and direction of bullets in shooting incidents.

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