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Conceptual Question on Impuldse of inelastic collision

  • Thread starter Tribean
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  • #1
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Homework Statement

If two objects are moving towards each other with two different velocities and two different mass then crash and stick together, find the force acted upon each object during the collision if the collision occurred over some time "t"



Homework Equations

m1v1+m2v2 = (m1+m2)v[itex]_{f}[/itex]

impulse = dp/dt



The Attempt at a Solution

I just have a basic question on the use of the mass and velocities within the impulse equation.

Would we use the final velocity and mass and subtract the initial velocities and mass of the objects. as so:

((m1+m2)(v_f) - (m1)(v1))/t
to get the impulse of the collision on object one? and do the same for object two.

OR

do we use their distinct momentum to subtract from one another:

((m1)(v1) - (m2)(v2))/t

I'm just having a hard time understanding the use of the equation conceptually. Can anyone help dumb it down a bit for me? Thanks!
 

Answers and Replies

  • #2
NascentOxygen
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Hi Tribean. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

After the collision they share a common velocity. Before the collision each body had its own mass and velocity so you know its pre-collision momentum. Therefore, for each body you can calculate its individual change in momentum, Δp.
 
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  • #3
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Hi, Nascent

Thank you for replying. So even though the objects are acting as one particle, I would use the added masses and the common velocity as the final momentum minus the pre - collision momentum of one of the objects:

((m1+m2)(v_f) - (m1)(v1))/t

or would I use its initial mass and new velocity?
 
  • #4
rude man
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What's the change in momentum before & after the collision of m1? Of m2?
 
  • #5
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Hi rudeman, sorry there are no actual variables for m1 or m2 or the momentum as I'm just asking for a conceptual view. I was just confused if we are to take both the masses combined together when taking the change in momentum

as I described in my last earlier response with the equation.
 
  • #6
NascentOxygen
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the objects are acting as one particle, I would use the added masses and the common velocity as the final momentum
Correct.

The pre-collision momentum of each is its mass x its velocity.
 
  • #7
rude man
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Hi rudeman, sorry there are no actual variables for m1 or m2 or the momentum as I'm just asking for a conceptual view. I was just confused if we are to take both the masses combined together when taking the change in momentum

as I described in my last earlier response with the equation.
I meant using m1, m2, v1 and v2 symbolically. It was a hint, not an inquiry.
 

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