Ballistic Pendulum kinetic energy

Click For Summary

Homework Help Overview

The problem involves a ballistic pendulum where a bullet is fired into a pendulum, and participants are tasked with calculating kinetic energy, momentum, and height after the collision. The subject area includes concepts from mechanics, specifically energy conservation and momentum.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of kinetic energy formulas and the conservation of momentum. There is an exploration of how kinetic energy is not conserved in inelastic collisions and the implications of energy loss due to various factors.

Discussion Status

Some participants have offered guidance on using conservation of momentum to find velocity after the collision. Others are questioning assumptions about energy conservation in different scenarios, such as when friction is neglected. The discussion reflects a range of interpretations regarding energy transfer and conservation principles.

Contextual Notes

Participants are navigating the constraints of the problem, including the physical limitations of the pendulum's length and the nature of the collision being inelastic. There is also mention of the effects of friction and energy dissipation in real-world scenarios.

QuarkCharmer
Messages
1,049
Reaction score
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:

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

??

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

There is no way this is right, the shaft of the pendulum is only 70cm long!
 
Physics news on Phys.org
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.
 
Would I use the conservation of momentum?

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

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

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

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

KE = \frac{1}{2}mv^{2}

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).
 
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.
 
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.
 
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.
 

Similar threads

Replies
7
Views
7K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
3
Views
2K
  • · Replies 29 ·
Replies
29
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
1
Views
2K
Replies
5
Views
1K