Solving a 2kg Ball Collision Problem

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Homework Help Overview

The discussion revolves around a collision problem involving two identical balls, each with a mass of 2.0 kg. The first ball, initially moving at a speed of 3.0 m/s, collides elastically with a stationary second ball. Participants are tasked with determining the speeds and directions of both balls post-collision, particularly focusing on the first ball's speed after the collision and the second ball's speed and direction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of momentum and energy conservation equations to analyze the collision. There are attempts to break down the problem into x and y components, with some questioning the implications of the identical masses of the balls. Others express confusion about how to relate the angles and components effectively.

Discussion Status

Several participants have provided insights into using conservation laws and have suggested combining equations to eliminate variables. There is active engagement with the mathematical relationships involved, though some participants express uncertainty about their progress and the complexity of the equations.

Contextual Notes

Participants note the challenge of having three unknowns while only having two equations, prompting discussions about the need for additional relationships, such as energy conservation. There is also mention of using trigonometric identities to simplify the equations.

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Homework Statement



A 2.0 kg ball moving with a speed of 3.0m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30 to the original path, determine
a) the speed of the first ball after the collision.
b) the speed and direction of the second ball after the collision

Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2


The Attempt at a Solution


X component:
v1 = v1' + v2' (since v2 is 0, and m's are equal)

Y component:
v1' = v2' (since v1 and v2 are 0 and m's are equal)

Because of this, does this mean that the angles should be equal as well?
I'm very confused about what I can deduce and assume from the identical mass of the balls..

It'd be great if someone could point me in the right direction. Thanks!
 
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Vx= Vcos theta and simailry Vy= Vsint heta can be used here.
 


So then do I divide the two components?
Cuz I tried that,

using
X: m1(v1 - v1'cos30) = m2(v2'costheta - 0)
v1 - v1'cos30 = v2costheta

Y: m1(0 -v1'sin30) = m2(v2'sintheta - 0)
- v1'sin30 = v2'sin theta

IF I divide Y by X, I get
v1-v1'sin30 / -v1'cos30 = tantheta

Then I'm stuck again :S
 


You have 3 unknowns but only 2 equations. You need to use conservation of energy as well.
 


Using conservation of energy,
v1 = v1'+v2'

with the combined X and Y component equations I've been trying to solve this and came up with several crazy dead-end solutions, using different trig identities..
 


kehwon said:
Using conservation of energy,
v1 = v1'+v2'
Try squares of velocities.
 


Okay,

so I have
(v1 - v1')(v1+v1') = v2'^2

which I would liek to incorporate into the x component: v1 - v1'cos30 = v2'costheta, but the cos30 is getting in the way.

I also tried putting Y component, with v2' = v1'sin30 / sin theta into the X component...
 


Your two momentum equations have in them sin theta and cos theta. Can you see how to combine them in a way that eliminates theta?
 


can I square the two momentum equations? then somehow combine them into sin^2theta + cos^2theta = 1?
 
  • #10


Yes!
 
  • #11


I'm having troubles linking the two. If I just add the two squard equations together, that makes sense since it would be like pythagorean theorem.

However, since theX componenti s v1 - v1'cos30 on the left side, square of that would be v1^2 - 2v1v1'cos30 + v1'^2...
 
  • #12


Sounds right. Why does that bother you? Now combine that with your energy equation and eliminate v2'
 
  • #13


great! thanks for all your help!
 

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