Conservation of Momentum problem

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Homework Help Overview

The discussion revolves around a conservation of momentum problem involving two blocks moving on a frictionless surface, with a focus on their velocities before and after a collision. The original poster attempts to calculate the final velocity of one block after the collision based on given masses and initial velocities.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster describes their method of applying the conservation of momentum equation but questions the validity of their results. Some participants suggest reconsidering the assumptions about the type of collision and the final velocities of the blocks.

Discussion Status

Participants are exploring different interpretations of the conservation of momentum, particularly distinguishing between elastic and inelastic collisions. Some guidance has been offered regarding the correct application of the momentum conservation principle, indicating a productive direction in the discussion.

Contextual Notes

The original poster's calculations are based on the assumption that the blocks stick together after the collision, which has been questioned by other participants. There is also a mention of the initial conditions changing when one block's velocity is negative.

Lannie
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I'm not sure what I've done wrong on this problem, so anyone who sees my mistake, please let me know!

Two blocks are moving on a frictionless surface. The first block's mass is 1.17 kg and its initial velocity is 5.50 m/s. The second block's mass is 2.73 kg and its initial velocity is 2.15 m/s.

What is the speed of the 1.17 kg block after the collision if the speed of the second block is 5.26 m/s after the collision?


(For clarification, before the collision the blocks are moving to the right, which I stated as the positive direction, and after the collision they are both still moving to the right.)

To solve, I used the equation Vfinal= (m1 x v1initial) + (m2 x v2initial) / (m1+m2)

so my work shows Vfinal = (1.17x5.50) + (2.73x2.15) / (1.17+2.73)
which worked out to approx. 3.16 m/s

But this approach didn't get me the right answer, which I don't understand since it seemed really straightforward.

The second part of the question asks,

If the initial velocity of the 2.73 kg block is the same magnitude but in the opposite direction, what is the velocity of the 1.17 kg block after the collision?

I used the exact same approach for this, but made the initial velocity of the second block negative. So,

Vfinal= (1.17x5.50) + (2.73x - 2.15) / (1.17+2.73)

which worked out to 0.145 m/s

this wasn't the right answer either.

Any help would be greatly appreciated!
 
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you must remember...that after the collision they do not stick together so their final velocity is not the same...remember how you have the velocity of the second afterwards...

it should be more like...
vf = [ (m1 x v1 initial) + (m2 v v2 initial) - (m2 x v2 final) ] / m1
 
You are confusing the general conservation law of momentum with the special case of INELASTIC COLLISION.
Remember: At the end of an inelastic collision, the two objects stick together with the same velocity.
That is what you've used, rather than the general law of conservation of momentum:
[tex]m_{1}v_{1,init.}+m_{2}v_{2,init.}=m_{1}v_{1,final}+m_{2}v_{2,final}[/tex]
(The notation ought to be self-explanatory)
Use this approach and see if you get better results..
 
actually yeah, WOW, I just looked at my work and was hoping to scramble here and edit this... totally missed a step there. I'm sorry! I'm a bonehead.
thanks though :)
 

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