Kinetic Energy Lost in Bullet Penetrating Can

  • Thread starter Thread starter Mivz18
  • Start date Start date
  • Tags Tags
    Bullet
AI Thread Summary
The discussion revolves around calculating kinetic energy lost when a bullet penetrates a pop can. The initial momentum of the bullet and can system is calculated as 5.4 kg·m/s, while the final momentum of the bullet is 3.6 kg·m/s. After the bullet emerges, the can's speed is determined to be approximately 32.73 m/s. The user initially struggles with calculating the kinetic energy lost, mistakenly setting up an equation that does not balance. Ultimately, the correct approach involves using the equation KE_f = KE_0 + ΔE to solve for the unknown energy loss.
Mivz18
Messages
44
Reaction score
0
Problem:

A bullet of mass 0.018 kg and initial speed 300 m/s penetrates an initially stationary pop can of mass 0.055 kg and emerges with a speed 200 m/s.

A) What is the initial momentum of the bullet and pop can system?
I found this answer by 0.018 * 300 = 5.6

B) What is the final momentum of the bullet?
I found this answer by 0.018 * 200 = 3.6

C) How fast is the can moving after the bullet emerges?
I found this answer by 0.018(300-200) and then took that quantity and divided it by 0.055 = 32.73 m/s

D) How much kinetic energy was lost (to heat, sound, deformation of can and bullet,...) in the process? Give your answer as a positive number.

Here is my problem. I don't know how to obtain this. I thought that maybe I could calculate it by KE1 + KE2 = KE1 + KE2 where the left side of the equation is before the collision and the right side is after. Then when I add the right side, it is a little off from the left, so I thought that was the KE lost. However, the online program I'm using says it isn't. What am I doing wrong or how can I go about achieving this?
 
Physics news on Phys.org
I think you have the right idea. Perhaps you missed something setting up your equation - can you show it with numerical values filled in?
 
This is what I got:

(1/2)m1v1 + (1/2)m2v2 = (1/2)m1v1 + (1/2)m2v2
(1/2)(0.018)(300^2) + (1/2)(0.055)(0^2) = (1/2)(0.018)(200^2) + (1/2)(0.055)(32.73^2)

From this I get 810 = 389.46
where 810 - 389.46 = 420.54 Lost ??
 
nevermind, answered my own question, lol. Thanks!
 
Yeah, you might want to set up the equation like this:
KE_f=KE_0+\Delta E
So that you end up solving for an unknown rather than setting up a broken equality.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top