Initial velocity of a bullet

  • #1

Homework Statement


A bullet of mass 12.0 grams is traveling with a speed V0 and strikes a block of mass M = 1.00 kg that rests on a horizontal surface (mu = 0.40). After the bullet imbeds in the block, it slides 2.60 meters before coming to rest. The initial velocity of the bullet must have been ? m/s.


Homework Equations


pi=pf
We were told that energy was not lost to anything except friction so
Ei+Wfriction=Ef


The Attempt at a Solution


Ei+Wfriction=Ef
(1/2)mvi2+μmgd=(1/2)mvf2
eliminated potential energy because no height is given
(1/)mvi2=μmgd
(1/2)(0.012kg)(vi2)=(.40)(1.012)(9.8)(2.60)
vi2=1719.050667
vi=41.461 m/s

Which isn't right. I think my error may lie in my assumption that the final KE is when the block has stopped and not right after impact, but I'm not certain.
 

Answers and Replies

  • #2
TSny
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Your calculation does not take into account the friction between the bullet and the block as the bullet embeds itself in the block. This causes additional loss of kinetic energy in the system during the collision ("inelastic collision"). Your calculation would be correct if Ei is the kinetic energy remaining just after the collision.

The kinetic energy just after the collision involves the velocity of the system just after the collision. See if you can relate this velocity to the initial velocity of the bullet using momentum concepts.
 
  • #3
Would that be m1v1=(m1+m2)v'? And then solving for v', and using that velocity in (1/2)mv^2 to find the KE immediately after the collision?
 
  • #4
tms
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Ei+Wfriction=Ef
Since the final energy is zero, the above implies that the initial energy is also zero.
Which isn't right. I think my error may lie in my assumption that the final KE is when the block has stopped and not right after impact, but I'm not certain.
If that were true, then there would have been no loss to friction.
 
  • #5
tms
644
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Your calculation does not take into account the friction between the bullet and the block as the bullet embeds itself in the block. This causes additional loss of kinetic energy in the system during the collision ("inelastic collision").
Since the problem gives no information from which such loss can be calculated, it must be assumed that the collision was elastic.
 
  • #6
Matterwave
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Since the problem gives no information from which such loss can be calculated, it must be assumed that the collision was elastic.
Since the bullet embeds itself in the block, the collision is perfectly INelastic - certainly not elastic.
 
  • #7
Matterwave
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Would that be m1v1=(m1+m2)v'? And then solving for v', and using that velocity in (1/2)mv^2 to find the KE immediately after the collision?
Yes, you have the right approach here.
 
  • #8
tms
644
17
Since the bullet embeds itself in the block, the collision is perfectly INelastic - certainly not elastic.
[strike]An elastic collision is one in which kinetic energy is conserved. While in real life there would certainly be some energy loss in this situation, for the purposes of an introductory physics problem one can assume either no loss or a negligible loss. Otherwise, how would you propose to calculate the loss in this case?[/strike]
Disregard.
 
Last edited:
  • #9
HallsofIvy
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No, Matterwave is correct- the bullet imbedding itself in the block is a "perfectlyinelastic" collision. You cannot use "conservation of energy", you must use "conservation of momentum" to determine the speed of the "block+ bullet" system immediately after the bullet lodges in the block. You can then use the kinetic energy of that system as the energy that must be lost due to friction.
 
  • #10
gneill
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An elastic collision is one in which kinetic energy is conserved. While in real life there would certainly be some energy loss in this situation, for the purposes of an introductory physics problem one can assume either no loss or a negligible loss. Otherwise, how would you propose to calculate the loss in this case?
It is an inelastic collision. Kinetic energy will not be conserved through the collision, therefore you don't bother with the KE on either side of the collision and apply conservation of momentum instead. Momentum is ALWAYS conserved.
 
  • #11
TSny
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Otherwise, how would you propose to calculate the loss in this case?
As already noted, you can use momentum conservation to relate the initial velocity of the bullet to the velocity of the bullet-block system immediately after the collision. Then you can find the loss of KE. In the case of a bullet embedding in a block, there will be a great loss of KE. For the numbers in this problem, you will discover that 98.8% of the kinetic energy of the bullet will be transformed to other forms of energy in the collision.
 
  • #12
Your calculation does not take into account the friction between the bullet and the block as the bullet embeds itself in the block. This causes additional loss of kinetic energy in the system during the collision ("inelastic collision"). Your calculation would be correct if Ei is the kinetic energy remaining just after the collision.

The kinetic energy just after the collision involves the velocity of the system just after the collision. See if you can relate this velocity to the initial velocity of the bullet using momentum concepts.
I went with this approach, found KE in terms of the initial velocity, set that equal to the work by friction, solved for the initial velocity and got the right answer. Thank you all.
 
  • #13
tms
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Disregard. I'm an idiot.
 
Last edited:
  • #14
tms
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Disregard. I'm an idiot.
 
Last edited:
  • #15
gneill
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Initial momentum is [itex]m_{bullet}v_0[/itex], and final momentum is zero. You have to assume an elastic collision in order to solve the problem.
No, the collision is inelastic. The assumption is that the bullet comes to rest with respect to the block in a very short time so that the block's motion with respect to the floor is insignificant during the actual collision. An essentially infinitesimal time after the collision the block and bullet have the same momentum as the bullet alone had before the collision. It's only after the block begins moving that friction couples the system to the floor (the previously effectively isolated system becomes an open one, or at least now includes the rest of the planet!).
 
  • #16
tms
644
17
No, the collision is inelastic. The assumption is that the bullet comes to rest with respect to the block in a very short time so that the block's motion with respect to the floor is insignificant during the actual collision. An essentially infinitesimal time after the collision the block and bullet have the same momentum as the bullet alone had before the collision. It's only after the block begins moving that friction couples the system to the floor (the previously effectively isolated system becomes an open one, or at least now includes the rest of the planet!).
I'm wrong. Someone must have toggled the ON/OFF switch in my brain.
 
  • #17
gneill
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I'm wrong. Someone must have toggled the ON/OFF switch in my brain.
No worries. Happens to everyone. Back on the horse! :smile:
 

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