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

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SUMMARY

This discussion centers on solving an AP Physics problem involving a collision between a 5-kilogram ball and a 0.5-kilogram cube. The cube, moving at 26 meters per second, strikes the ball, which then leaves the table. The participants determined the horizontal velocity of the ball after the collision to be 4.727 m/s to the right and the cube's speed to be 21.27 m/s to the left. The collision is confirmed to be inelastic, as kinetic energy is not conserved, despite momentum being conserved.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with impulse and its calculation from force-time graphs
  • Knowledge of elastic vs. inelastic collisions
  • Ability to solve equations involving mass, velocity, and momentum
NEXT STEPS
  • Study the concept of impulse and how to calculate it from force-time graphs
  • Learn about the differences between elastic and inelastic collisions
  • Explore momentum conservation in two-dimensional collisions
  • Practice solving AP Physics problems involving collisions and impulse
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AP Physics students, educators teaching mechanics, and anyone interested in understanding collision dynamics and momentum conservation.

  • #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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