Physics AP problem (mechanics)

  • #26
I don't understand your abbreviations---what does "I" stand for, momentum?

So once you have the following, you only know v(initial) so how do you solve because there are two unknowns----v1 and v2, right? If you can find v1 or v2, how do you do this without making an energy equation?
I(initial) = I(final)
.5v(initial) = .5(v1) + 5(v2)
 
  • #27
physicsgirl101 said:
I don't understand your abbreviations---what does "I" stand for, momentum?

So once you have the following, you only know v(initial) so how do you solve because there are two unknowns----v1 and v2, right? If you can find v1 or v2, how do you do this without making an energy equation?
I(initial) = I(final)
.5v(initial) = .5(v1) + 5(v2)

yes I = momentum

The graph is the key... As Mason had already said... the area under the graph is the impulse acted on the ball... and lost by the cube...
 
  • #28
"The graph is the key... As Mason had already said... the area under the graph is the impulse acted on the ball... and lost by the cube..."

The only thing I previously knew about momentum was that momentum was Force x Time (so it made sense that it was the area under the graph, which is 12. And I understand that that means 12 is the impulse acted on the ball and lost by the cube---but I don't know how I can use that information..... I don't know any other equations that involve impulse and I need a way of finding v1 or v2 so that I can solve for the other one..... any more help would be greatly apprectiated!
 
  • #29
physicsgirl101 said:
"The graph is the key... As Mason had already said... the area under the graph is the impulse acted on the ball... and lost by the cube..."

The only thing I previously knew about momentum was that momentum was Force x Time (so it made sense that it was the area under the graph, which is 12. And I understand that that means 12 is the impulse acted on the ball and lost by the cube---but I don't know how I can use that information..... I don't know any other equations that involve impulse and I need a way of finding v1 or v2 so that I can solve for the other one..... any more help would be greatly apprectiated!

Impulse = change in momentum.... I'm sure you've seen this before... you had to if you are in physics AP... cuz i have....
 
  • #30
I haven't really seen it before because we just started with momentum on friday and all my professor said was impulse = time x force......
 
  • #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?
 
  • #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 cant believe i didnt catch that..... i think im 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, im glad to be done, thanks for all the clarifications!
 

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