AP Problem - Elastic Collision at an Angle

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Homework Help Overview

The discussion revolves around an elastic collision problem involving two masses on a frictionless surface, with a focus on the conservation of momentum and kinetic energy principles. Participants are analyzing the effects of the collision at an angle and the implications of the setup on the velocities post-collision.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation laws relevant to the problem, questioning how the angle of collision affects the velocities. There is also an exploration of the relationship between initial and final velocities, particularly in the context of kinematic equations.

Discussion Status

Some participants are clarifying the implications of a frictionless surface on the collision dynamics. There is an ongoing examination of the correctness of calculated values and the reasoning behind discrepancies between personal calculations and provided solutions.

Contextual Notes

Participants are working with specific values and conditions from the problem statement, including a diagram referenced in the discussion. There is mention of the AP board's expected answers, which introduces a potential constraint on the interpretations of the problem.

meganw
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Homework Statement



SEE BELOW: 4th Reply has a Diagram

Homework Equations



Conservation of Momentum: m1(v1i) +m2(v2i) = m1(v1f) + m2(v2f)
Vf^2=Vi^2 + 2a(delta y)
Conservation of Kinetic Energy (Elastic Collision): .5m1(v1i^2)+.5m2(v2i) = .5m1(v1f^2)+.5m2(v2f^2)

The Attempt at a Solution



See 4th post for newest question:
 
Last edited:
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Since the surface is frictionless, only the component of velocity normal to the surface is affected by the collision.
 
What? Sorry I don't understand what you're saying...does that mean the angle doesn't change?

(By the way thank you for being so amazingly helpful! :) )
 
Yes, the angle the velocity makes with the surface will not change.
 
This is the problem and diagram:

http://img255.imageshack.us/img255/1966/55320227sg8.png

I have done a-c, and these are my answers that I got. I know they're correct because I checked them with the solutions:

L= 4\sqrt{}2 (h)
Delta Y = Delta x = L/\sqrt{}2

For d I used the kinematic equation
Vf^2=Vi^2 + 2a(delta y)
I got delta y=4h
and vf = \sqrt{}8gh

but the ap board says the answer is (conservation of energy):

mgh + mgL/\sqrt{}2=.5mv^2

v = \sqrt{}10gh

But why is my answer for part d wrong?

note: \sqrt{}2
(this symbol is the square root symbol)
 
Last edited by a moderator:
gahhhh
 
You didn't add the Vi^2 (=2gh) to the 2a(delta y) (=8gh).
 

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