# Final velocities of two objects in a 2D elastic collision

Protium_H1

## Homework Statement

An atomic nucleous of mass m traveling with speed v collides elastically with a target particle of mass 3.0m (initially at rest) and is scattered at 45o
(a). What are the final speeds of the two particles?
Advice: eliminate the target particle's recoil angle by manipulating the equation of momentum conservation so that can use the identity sin2Φ + cos2Φ = 1

Variables (that I used):

v = speed of the atom before the collision (along the x-axis, i.e. parallel to the x-axis)
u1 = final velocity of the atom after the collision (goes above the x-axis by 45o)
u2 = final velocity of the target particle after the collision (goes below the x-axis by Φ)

## Homework Equations

[/B]
Conservation of Momentum:
m1v1 + m2v2 = m1u1 m2u2

## The Attempt at a Solution

[/B]
I used the conservation of momentum formula and simplified it as follows:

m1v1 + m2v2 = m1u1 + m2u2
mv = mu1 + 3.0mu2
mv = m(u1 + 3.0u2)
v = u1 + 3.0u2
I broke u1 and u2 into component forms:

x-direction:
v = u1cos(45o) + 3.0u2cosΦ
y-direction:
0 = u1sin(45o) - 3.0u2sinΦ
This is where I get stuck, I even tried to use the conservation of kinetic energy,

v2 = u12 + 3.0u22
To find for the unknowns, but for some reason, the equations just all cancel out or become impossible to solve when I tried to substitute and find the final velocities, and for some reason, I can't get the expression sin2Φ + cos2Φ = 1 as the Advice recommended me to do. I asked my Prof. and he said that I had to break the final velocities into its component forms to get that trig. identity before rushing to a meeting.

Note:

Please correct me if I'm wrong but, I think that the angle between these two final velocities (45o + Φ) = 90o, since the conservation of kinetic energy equation: v2 = u12 + 3.0u22, looks like Pythagoras Theorem, where v is the hypotenuse, and all these 3 velocities form a right triangle.

Homework Helper
Gold Member
Welcome to PF, Protium_H1.

x-direction:
v = u1cos(45o) + 3.0u2cosΦ
y-direction:
0 = u1sin(45o) - 3.0u2sinΦ
This is where I get stuck, I even tried to use the conservation of kinetic energy,

v2 = u12 + 3.0u22
To make use of the trig identity, solve the x-momentum equation for 3.0u2cosΦ and solve the y-momentum equation for 3.0u2sinΦ. Can you then see how to proceed?

Note:

Please correct me if I'm wrong but, I think that the angle between these two final velocities (45o + Φ) = 90o
This would be true if the two masses were equal. It is not true in this problem.

Protium_H1
Welcome to PF, Protium_H1.

To make use of the trig identity, solve the x-momentum equation for 3.0u2cosΦ and solve the y-momentum equation for 3.0u2sinΦ. Can you then see how to proceed?

This would be true if the two masses were equal. It is not true in this problem.
Welcome to PF, Protium_H1.

To make use of the trig identity, solve the x-momentum equation for 3.0u2cosΦ and solve the y-momentum equation for 3.0u2sinΦ. Can you then see how to proceed?

This would be true if the two masses were equal. It is not true in this problem.

Would I have to add these two equations up and square them?

[3.0u2cosΦ]2 = [v - u2cos(45o)]2
[3.0u2sinΦ]2 = [u1sin(45o)]2

then,

[3.0u2sinΦ]2 + [3.0u2cosΦ]2 = [u1sin(45o)]2 + [v - u2cos(45o)]2
9.0u22[cos2Φ + sin2Φ] = [u1sin(45o)]2 + [v - u2cos(45o)]2
9.0u22 = [u1sin(45o)]2 + [v - u2cos(45o)]2
but I still have 3 unknowns in this equation, is there a way to cancel 2 of them?

Homework Helper
Gold Member
Would I have to add these two equations up and square them?

[3.0u2cosΦ]2 = [v - u2cos(45o)]2
[3.0u2sinΦ]2 = [u1sin(45o)]2
Yes. But you have a typographical error in the subscript on the right side of the first equation.

but I still have 3 unknowns in this equation, is there a way to cancel 2 of them?
You have only two unknowns since you can consider v as given.
Don't forget the energy equation.