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 45^o
(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 45^o)
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(45^o) + 3.0u₂cosΦ
y-direction:
0 = u₁sin(45^o) - 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 (45^o + Φ) = 90^o, 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.

​

 

[TSny]

Welcome to PF, Protium_H1.

x-direction:
v = u₁cos(45^o) + 3.0u₂cosΦ
y-direction:
0 = u₁sin(45^o) - 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 (45^o + Φ) = 90^o

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(45^o)]²
[3.0u₂sinΦ]² = [u₁sin(45^o)]²​

then,

[3.0u₂sinΦ]² + [3.0u₂cosΦ]² = [u₁sin(45^o)]² + [v - u₂cos(45^o)]²
9.0u₂²[cos²Φ + sin²Φ] = [u₁sin(45^o)]² + [v - u₂cos(45^o)]²
9.0u₂² = [u₁sin(45^o)]² + [v - u₂cos(45^o)]²
​

but I still have 3 unknowns in this equation, is there a way to cancel 2 of them?

 

[TSny]

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

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