Momentum and energy in collisions, bullet problem

AI Thread Summary
A bullet of mass 0.125 kg traveling at 250 m/s embeds into a block of mass 3.5 kg, resulting in an inelastic collision. The final speed of the block and bullet together can be calculated using conservation of momentum. The kinetic energy before the collision is higher than after, with the difference indicating energy lost to thermal energy. The rise in thermal energy can be determined by subtracting the final kinetic energy from the initial kinetic energy. The discussion also highlights confusion regarding the correct formulas for kinetic energy and momentum in inelastic collisions.
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Homework Statement



A bullet of mass 0.125 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface.
(a) What is the speed of the block after the bullet embeds itself in the block?

v = ? m/s

(b) Calculate the kinetic energy of the bullet plus the block before the collision:

Ki = ? J

(c) Calculate the kinetic energy of the bullet plus the block after the collision:

Kf = ? J

d) Was this collision elastic or inelastic?
inelastic



(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:

Ethermal,bullet + Ethermal,block = ? J

(f) Which of the following statements are true?
The bullet and block got hot because Q was large.
Because the process was very fast, Q was negligible.
Q was negative during the collision because the bullet and block were hotter than the surroundings.


Homework Equations


delta p1+delta p2=0
delta K=0


The Attempt at a Solution


For a.) i tried 2(m/M)v but it was wrong.
 
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jchojnac said:

Homework Statement



A bullet of mass 0.125 kg traveling horizontally at a speed of 250 m/s embeds itself in a block of mass 3.5 kg that is sitting at rest on a nearly frictionless surface.
(a) What is the speed of the block after the bullet embeds itself in the block?
v = ? m/s
(b) Calculate the kinetic energy of the bullet plus the block before the collision:
Ki = ? J
(c) Calculate the kinetic energy of the bullet plus the block after the collision:
Kf = ? J
d) Was this collision elastic or inelastic?
inelastic

(e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision:

Ethermal,bullet + Ethermal,block = ? J

(f) Which of the following statements are true?
The bullet and block got hot because Q was large.
Because the process was very fast, Q was negligible.
Q was negative during the collision because the bullet and block were hotter than the surroundings.

Homework Equations


delta p1+delta p2=0
delta K=0

The Attempt at a Solution


For a.) i tried 2(m/M)v but it was wrong.

What is the equation for an inelastic collision then?

Does this help?
http://en.wikipedia.org/wiki/Inelastic_collision#Perfectly_inelastic_collision
 
yea thanks
 
for c.) wouldn't the final kinetic energy be 1/2*m*v^2 of just the block since the final speed of the bullet is 0?
 
jchojnac said:
for c.) wouldn't the final kinetic energy be 1/2*m*v^2 of just the block since the final speed of the bullet is 0?

In an inelastic collision the bullet will have the same velocity as the block. It is embedded isn't it? So the final kinetic energy will be the sum of the masses times the final velocity.
 
i tried (m1+m2)*vf...(.125+3.5)*8.62 and got 31.2475. It said it was wrong. I calculated the final velocity and got it right. What's wrong?
 
jchojnac said:
i tried (m1+m2)*vf...(.125+3.5)*8.62 and got 31.2475. It said it was wrong. I calculated the final velocity and got it right. What's wrong?

Try 1/2 *(m1+m2)*vf^2...this should work
 
jchojnac said:
i tried (m1+m2)*vf...(.125+3.5)*8.62 and got 31.2475. It said it was wrong. I calculated the final velocity and got it right. What's wrong?

Isn't c) asking for 1/2 m*v2?

1/2*3.65*(8.62)2
 
For part e) take the difference between the kinetic energy of the bullet plus the block before the collision and the kinetic energy of the bullet plus the block after the collision to get the answer.
f) it happens so fast, that q is negligible.
 
  • #10
yea i got it now
 
  • #11
hey, are you in 172??
is this on the webassign?
what did you get for the first problem? the one with different masses? i have a posting on here, i tried many different things but can't get it.
can you help me on it??
 
  • #12
Yea man, I am in 172 and on webassign, I am stuck at the same spot you are.
 
  • #13
problem one answers are

|1,final| < |1,initial|

If m2 >> m1, then the final speed of object 1 is greater than the final speed of object 2.

1,initial = 1,final + 2,final

is that where you were stuck?
i guess we had to assume it was elastic where M1 bounces off of M2.
not sure how the expect us to know that with what is given but oh well.
 
  • #14
No, sorry I am mean on question #3 where you got stuck...I got #1
 
  • #15
Question #3 i found the Vfinal by using
P1(inital) + P2(inital) = P1(final) + P2(final)
you know all but P2(final) just solve for it by doing m*v (find the P) for each
remember your answer is in P so use P= mv to get v (just divide by mass)

not sure how to find E internal. I am still stuck on that.

lastly the box at the bottom just click all the boxes with Q = 0 (should be 2 boxes)
 
  • #16
alrite man thanks!
 
  • #17
the only thing I am stuck on is the e internal
 
  • #18
no problem.
see you on here tomorrow or thursday for the homework due then lol
its worth 67 points but not sure how hard it is.
good luck on it lol
 
  • #19
yeah i did not get that. got everything else so 21/22 is fine with me
 
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