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Perfectly inelastic collisions: Bullet hitting a block: Force, velocity and distance?

  • Thread starter Merlinnair
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  • #1

Homework Statement


A 50 g bullet at 50 m/s hits a block 2.0 m away. The coefficient of friction is 0.30. What is the force of bullet on the block, the velocity of the block and bullet and distance traveled by the block? Assume this is a perfectly inelastic collision.

Known: m_b = .0050kg
v_b = 50. m/s
d_b = 2.0 m
μ = 0.30


Homework Equations


p_1 = p_2
F = ma
μ = -a/g
Fd=1/2mv^2

The Attempt at a Solution


A) 0.30 = -a/9.8

I feel like this isn't right for some reason...would this give me the correct acceleration?


B) (.0050kg)(50 m/s)=(m_1 + m_2)v

Since it's perfectly inelastic, can't I assume that v is the same for the block? Is that how I'd find m2?

C) I would be able to solve for distance once I have the velocity.
 

Answers and Replies

  • #2
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I think you should consider the coefficient of restitution and what it means in this situation which is a perfectly inelastic collision with friction.
 
  • #3
gneill
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The question statement seems incomplete to me. Shouldn't we be given the mass of the block? And "the force of the bullet on the block" has no meaning without knowing the depth of penetration or the time for the bullet to come to a halt within the block.
 
  • #4
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I don't think you need to know the mass of the block, my following explanation may give away too much. In a perfectly inelastic collision, coefficient of restitution is zero. Also momentum is only conserved if there is no friction. There is friction here. I only stated basic principles associated with this problem, but I believe it is enough.
 
  • #5
gneill
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20,792
2,770


I don't think you need to know the mass of the block, my following explanation may give away too much. In a perfectly inelastic collision, coefficient of restitution is zero. Also momentum is only conserved if there is no friction. There is friction here. I only stated basic principles associated with this problem, but I believe it is enough.
I am looking forward to your elaboration of the solution with great anticipation :smile:
 
  • #6


Sorry, I may be forgetting a variable, as I don't have the question with me. Do I need the acceleration, or can I find the force by solving for kinetic energy?

And, I don't know what the coefficient of restitution is...

If m_2 is one of the given variables, then would my work for finding velocity be correct?

To find distance I was going to use 2ad = vf2 - vi2

But I still have the issue of finding acceleration.

And assuming I do have the mass of the block, how do I find the force of the bullet on the block?
 
  • #7
gneill
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20,792
2,770


Sorry, I may be forgetting a variable, as I don't have the question with me. Do I need the acceleration, or can I find the force by solving for kinetic energy?

And, I don't know what the coefficient of restitution is...

If m_2 is one of the given variables, then would my work for finding velocity be correct? If so, then I should be able to find the distance myself...

F = ma = -f = -μmg

So a = -μg
And then I plug a back into ma to find force....right?
If you know the mass of the block then you can find the velocity of the bullet+block the instant after the bullet becomes lodged in the block by conservation of momentum. After that the block and bullet move together as a single object, sliding with friction on the surface. The usual accelerated motion techniques then apply.

As for the force of the bullet on the block, I think you'll still need to know something else about the particulars of the collision, such as the depth of penetration or the time for the bullet to come to a stop within the block.
 
  • #8


As for the force of the bullet on the block, I think you'll still need to know something else about the particulars of the collision, such as the depth of penetration or the time for the bullet to come to a stop within the block.
I'm in a pretty low level physics class, I'm not even sure how to use them if I had those variables. Could I simply neglect that fact that the bullet embeds itself?
 
  • #9
gneill
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20,792
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I'm in a pretty low level physics class, I'm not even sure how to use them if I had those variables. Could I simply neglect that fact that the bullet embeds itself?
Part of the problem statement was to find the "force of bullet on the block". I don't know how you can do that without knowing some details about the collision. Knowing, for example, the initial and final velocities for the bullet and either the distance or time over which that velocity change happened, you can determine the average acceleration that the bullet experienced. That would give you the force (F = M*A).
 

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