How to Determine Velocities After a Collision in a Physics AP Mechanics Problem?

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The discussion revolves around solving an AP Physics problem involving a collision between a cube and a ball on a frictionless table. Participants are trying to determine the horizontal velocity of the ball after the collision and the speed and direction of the cube post-collision. The impulse graph provided is crucial for calculating the change in momentum, which is used to find the velocities. There is confusion about whether the collision is elastic or inelastic, with participants clarifying that it is inelastic due to the loss of kinetic energy. Ultimately, the area under the impulse graph represents the momentum transferred to the ball and lost from the cube, guiding the calculations for the final velocities.
  • #31
Um... the one problem i still have is everyone is saying equations that i do understand, but there are always two unknowns (v1 and v2) and someone said I already knew one---but I don't. So my real problem is finding one of them so I can find the other... If I have to use 2 equations and solve, what two equations can I set up?
 
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  • #32
Part d of the problem says: Determine the kinetic energy dissipated in the collision.

I know kinetic energy was lost, because it was an inelastic collision, but when I do it out I get that no kinetic energy is lost:

before collision:
(1/2)(.5)(26^2)+(1/2)(5)(0) = 169

after collision:
(1/2)(.5)(21.27^2) + (1/2)(5)(4.727^2) = approx. 169

I'm assuming I got this incorrect result because the velocities I used (21.27 and 4.727) are incorrect---because the collision would have to be elastic for them to be correct, and thus it makes sense that in an elastic collision the energy would be conserved... So basically this goes back to my last post----I need to find the correct v1 and v2... Can anybody help? Please!
 
  • #33
physicsgirl101 said:
Part d of the problem says: Determine the kinetic energy dissipated in the collision.

I know kinetic energy was lost, because it was an inelastic collision, but when I do it out I get that no kinetic energy is lost:

before collision:
(1/2)(.5)(26^2)+(1/2)(5)(0) = 169

after collision:
(1/2)(.5)(21.27^2) + (1/2)(5)(4.727^2) = approx. 169

I'm assuming I got this incorrect result because the velocities I used (21.27 and 4.727) are incorrect---because the collision would have to be elastic for them to be correct, and thus it makes sense that in an elastic collision the energy would be conserved... So basically this goes back to my last post----I need to find the correct v1 and v2... Can anybody help? Please!

Ok, you know F*t = m(Vf - Vi)

you already know F*t = 12 from the graph... and you know the ball was initially at rest... You also know the mass of the ball... I'm sure you can solve for vf, which gives you one of the velocities of the 2 objects...
 
  • #34
Thank you so much, APchemstudent! i can't believe i didnt catch that... i think I am in overload... :) THANKS!
 
  • #35
Alright... After a lot of work, this is what I got:
immediately after the collision:
speed of ball: 2.4m/s
speed of cube: 2m/s (going to the right---the same direction as the ball*)

I then gathered that 153.6 Joules of kinetic energy was lost in the collision...

How does that all sound? (especially the direction of the ball??)
 
  • #36
physicsgirl101 said:
Alright... After a lot of work, this is what I got:
immediately after the collision:
speed of ball: 2.4m/s
speed of cube: 2m/s (going to the right---the same direction as the ball*)

I then gathered that 153.6 Joules of kinetic energy was lost in the collision...

How does that all sound? (especially the direction of the ball??)

everything looks perfect...
 
  • #37
Yay! I'm so excited, thank you, everyone!
 
  • #38
there was another part of the question which asks the distance between the two spots of initial impact on the ground (cube and ball)... if anyone cares to do it... i got 4.178 meters-----anyways, I am glad to be done, thanks for all the clarifications!
 

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