Calculating Final Velocities of Elastic Collision in 2D

In summary, the Homework Statement is a conversation about homework and collision. A 1.2kg mass moving 2.4m/s [East] has an elastic collision with a 2.4kg mass moving 1.8m/s [North]. The final velocities are v1' = 2.3m/s [W38N] and v2' = 2.4m/s [E27N].
  • #1
Auras
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0

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?

-------------------------------------

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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  • #2
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.
 
  • #3
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)​
 
  • #4
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!
 

Related to Calculating Final Velocities of Elastic Collision in 2D

1. How do you calculate the final velocities of an elastic collision in 2D?

The final velocities in a 2D elastic collision can be calculated using the conservation of momentum and conservation of kinetic energy equations. The equations are:
m1v1i + m2v2i = m1v1f + m2v2f (conservation of momentum)
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2v2f^2 (conservation of kinetic energy)
where m1 and m2 are the masses of the objects, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.

2. What is an elastic collision?

An elastic collision is a type of collision between two objects where both the momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before and after the collision are the same. In an elastic collision, the objects involved do not stick together or deform upon impact.

3. What is the difference between elastic and inelastic collisions?

In an elastic collision, both the momentum and kinetic energy are conserved. However, in an inelastic collision, only the momentum is conserved while some kinetic energy is lost. Inelastic collisions typically involve objects that stick together or deform upon impact.

4. Can the final velocities of an elastic collision in 2D be negative?

Yes, the final velocities in a 2D elastic collision can be negative. This occurs when one object is moving in the opposite direction of the other object after the collision. The negative sign indicates that the object is moving in the opposite direction of the positive direction chosen for the coordinate system.

5. What happens if the masses of the objects in an elastic collision are different?

If the masses of the objects in an elastic collision are different, the final velocities will also be different. The object with the smaller mass will have a larger final velocity, while the object with the larger mass will have a smaller final velocity. This can be seen in the conservation of momentum equation, where the masses are inversely proportional to the velocities.

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