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2d momentum of identical without final velocities, but given angle of first object

  • Thread starter kehwon
  • Start date
1. The problem statement, all variables and given/known data

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

2. Relevant equations
m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2


3. 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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Re: 2d momentum of identical without final velocities, but given angle of first objec

Vx= Vcos theta and simailry Vy= Vsint heta can be used here.
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

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
 

haruspex

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Re: 2d momentum of identical without final velocities, but given angle of first objec

You have 3 unknowns but only 2 equations. You need to use conservation of energy as well.
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

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..
 

haruspex

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Re: 2d momentum of identical without final velocities, but given angle of first objec

Using conservation of energy,
v1 = v1'+v2'
Try squares of velocities.
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

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...
 

haruspex

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Re: 2d momentum of identical without final velocities, but given angle of first objec

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?
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

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

haruspex

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Re: 2d momentum of identical without final velocities, but given angle of first objec

Yes!
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

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...
 

haruspex

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Re: 2d momentum of identical without final velocities, but given angle of first objec

Sounds right. Why does that bother you? Now combine that with your energy equation and eliminate v2'
 
Re: 2d momentum of identical without final velocities, but given angle of first objec

great! thanks for all your help!
 

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