Momentum and energy in collisions, bullet problem

In summary, 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. The speed of the block after the bullet embeds itself in the block is v. The kinetic energy of the bullet plus the block before the collision is Ki. The kinetic energy of the bullet plus the block after the collision is Kf. The collision was inelastic. The rise in thermal energy of the bullet plus block as a result of the collision is Ethermal,bullet + Ethermal,block = ? J. The bullet and block got hot because Q was large. The equation for an inel
  • #1
jchojnac
39
0

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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  • #2
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
 
  • #3
yea thanks
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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
 
  • #8
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
 
  • #9
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
 

1. What is the difference between momentum and energy in collisions?

Momentum and energy are related but distinct concepts in collisions. Momentum is a measure of an object's mass and velocity, while energy is a measure of its ability to do work. In collisions, momentum is always conserved, meaning the total momentum before and after the collision remains the same. However, energy is not always conserved as it can be converted into other forms, such as heat or sound.

2. How does the mass and velocity of a bullet affect its momentum?

The momentum of a bullet is directly proportional to its mass and velocity. This means that a heavier bullet or a faster bullet will have a greater momentum compared to a lighter or slower bullet. This is why a larger bullet can cause more damage upon impact.

3. Can a bullet lose energy during a collision?

Yes, a bullet can lose energy during a collision. This can happen if the bullet hits an object that is able to absorb some of its energy, such as a soft target or a bulletproof vest. In these cases, the bullet will still retain its momentum but its energy will decrease due to the work done in overcoming the resistance of the target.

4. How does the conservation of momentum apply to a bullet problem?

In a bullet problem, the conservation of momentum applies to the total momentum of the bullet and the object it collides with. This means that the total momentum before the collision will be equal to the total momentum after the collision. This principle can be used to calculate the velocity or mass of either the bullet or the object after the collision.

5. Is there a relationship between the speed of a bullet and its kinetic energy?

Yes, there is a direct relationship between the speed of a bullet and its kinetic energy. The kinetic energy of an object is equal to half its mass multiplied by its velocity squared. This means that a bullet with a higher speed will have a greater kinetic energy compared to a bullet with a lower speed, assuming they have the same mass.

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