Calculating Final Velocities of Elastic Collision in 2D

AI Thread Summary
The discussion focuses on calculating the final velocities of two masses after an elastic collision in 2D. The initial attempt involved using conservation of momentum and kinetic energy equations, but confusion arose regarding the distinction between velocity components and total velocities. It was clarified that the problem lacks a unique solution due to having four unknowns but only three equations, indicating additional information is needed. The participant confirmed that the conservation of energy should account for both the North-South and East-West directions collectively. Ultimately, the problem was deemed unsolvable without further data, highlighting the importance of understanding the components involved in such calculations.
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Homework Statement



A 1.2kg mass moving 2.4m/s [East] has an elastic collision with a 2.4kg mass moving 1.8m/s [North]. What are the final velocities?

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I know that I need Ek before and after the collision, but I'm not sure how to calculate it for this scenario (a way that would allow me to substitute into Conservation of Momentum find velocities of both objects). I know it's a scaler quantity, so does that mean I do not have to consider making the energy perpendicular to each other?

Homework Equations



p = m1v1 + m2v2
Ek = 1/2 mv^2

The Attempt at a Solution



Here's what I've got so far but I don't know if it's right so I don't want to be doing repeated mistakes throughout my homework and then later on the unit test.

For E-W direction:
p = p'
m1v1 + m2v2 = m1(v1') + m2(v2')
1.2(2.4) + 0 = 1.2(v1') + 2.4(v2')
2.4 = v1' + 2(v2)'

Since it is an elastic collision, I attempted with pure energy since it's scaler, this is what I'm not sure about:
Ek = Ek'
1/2 m(v1)^2 + 1/2 m(v2)^2 = 1/2 m(v1')^2 + 1/2 m(v2')^2
1.2(2.4)^2 + 2.4(1.8)^2 = 1.2(v1')^2 + 2.4(v2')^2
14.688 = 1.2(v1')^2 + 2.4(v2')^2
12.24 = (v1')^2 + 2(v2')^2
12.24 - 2(v2')^2 = (v1')^2
Sqrt {12.24 - 2(v2')^2} = v1'

Then I substituted it into the momentum equation
2.4 = Sqrt {12.24 - 2(v2')^2} +2(v2)'

After squaring both sides and getting the roots of the quadratic equation, I got:
v2' = 2.1 or v2'= -0.51

I reasoned that the ball heading North (v2) should not suddenly start going West when the ball hitting it is heading East.

I end up with final E-W velocities of:
v2' = 2.1m/s [E] and v1' = 1.8m/s [W]
However, these are only Velocities in a East/West direction.

Am I doing this right?
I don't want to do a whole page of calculations for NS direction as well and then make a vector triangle just to find out I did the whole thing wrong.Thank you for the help!
 
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I went through with this since I've got no replies :(

Anyways, my results ended up to be:

v1' = 2.3m/s [W38N]
v2' = 2.4m/s [E27N]

These numbers don't seem to make sense at all... The heavier ball starts moving faster after collision and is directed more towards East than what it was initially traveling at in, which was North.

Could someone explain to me what I did wrong? I'm pretty sure my Kinetic Energy equation is iffy.
 
Welcome to PF :smile:

I have 2 comments:

Comment #1

The problem does not have a unique solution. We have 4 unknowns (v1x', v1y', v2x', v2y') but only 3 equations (momentum in x-direction, momentum in y-direction, and conservation of energy).

Another piece of information is required to solve the problem. Did the problem statement say anything else?

Comment #2

It seems you are not distinguishing between velocities, and velocity components, properly. For example, in the East-West momentum equation, the final "velocities" are really the x-components of the final velocities, i.e. v1x' and v2x'.

For the kinetic energy treatment, you got the correct initial energy of (1/2)14.688 J. The collision is elastic, as you said, so:

14.688 J = m1(v1x'2+v1y'2) + m2(v2x'2+v2y'2)​
 
Redbelly98 said:
Welcome to PF :smile:

I have 2 comments:

Comment #1

The problem does not have a unique solution. We have 4 unknowns (v1x', v1y', v2x', v2y') but only 3 equations (momentum in x-direction, momentum in y-direction, and conservation of energy).

Another piece of information is required to solve the problem. Did the problem statement say anything else?

Comment #2

It seems you are not distinguishing between velocities, and velocity components, properly. For example, in the East-West momentum equation, the final "velocities" are really the x-components of the final velocities, i.e. v1x' and v2x'.

For the kinetic energy treatment, you got the correct initial energy of (1/2)14.688 J. The collision is elastic, as you said, so:

14.688 J = m1(v1x'2+v1y'2) + m2(v2x'2+v2y'2)​

Thanks for your reply!

Oh, for the components, yes I understand what you mean, but the style that my teacher taught us was that he just subtitled each section so in the end as long as we make a vector triangle for the final velocity (sorry, I guess I should've specified components in each of the equations under each subheadings as well). So basically all the velocity values under the E-W subheading are for for the E-W component, or x-plane as you stated.

Also, it turned out that the problem was unsolvable. I asked my teacher the next day and there's stuff missing (angles after or any more data about the result) and that conservation of energy is the total energy for BOTH N-S and E-W direction, so the Ek equation I applied throughout the components wouldn't have given me the correct answer.

Thanks again!
 
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