# Solve for 2D Elastic Collision: Find θ & φ Angles

• amanda.ka
In summary: Correct the sign of sinθ in (2-cosθ)2 and solve for cosθ.1.341cosφ = 2-cosθcosθ = -1.341sinφThen you can use the trig identity cos2θ + sin2θ = 1 to solve for cosφ.cosφ = (-1.341)2cosθin summary,In summary, the final speed of puck 2 is 2 m/s and the angles θ and φ are θ = -1.341 and φ = 1.341.
amanda.ka

## Homework Statement

A 2D elastic collision:
Two pucks (masses m1 = 0.5 kg and m2 = 0.3 kg) collide on a frictionless air-hockey table. Puck 1 has an initial velocity of 4 m/s in the positive x direction and a final velocity of 2 m/s in an unknown direction, θ. Puck 2 is initially at rest. Find the final speed of puck 2 and the angles θ and φ.

I get stuck at the end of this problem when I have to use the two equations to solve for 2 unknown angles. If someone could show me how to do that last step that would be great. Thanks in advance!

## The Attempt at a Solution

Since the collision is elastic I found the final kinetic energy using Ei = Ef and it equals 4.47 m/s

Then conservation of the x and y components of total momentum:

X DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
(0.5)(4) = (0.5)(2cosθ) + (0.3)(4.47cosφ)
2 = cosθ + 1.341cosφ

Y DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
0 = (O.5)(2sinθ) -(0.3)(4.47sinφ)
0 = sinθ - 1.341sinφ

Hint: cos2θ + sin2θ = 1

(Your work looks good so far.)

TSny said:
Hint: cos2θ + sin2θ = 1

(Your work looks good so far.)
So I solve the first equation for cosφ and the second equation for sinφ?
1.341cosφ = 2-cosθ
and
-1.341sinφ = sinθ

then square each term:
1.3412cos2φ = 22-cos2θ
and
-1.3412sin2φ = sin2θ

then I added them but I'm still a bit lost/don't know the next step :/

1.3412cos2φ + (-1.341)2sin2φ = 22-cos2θ + sin2θ

Take another look at (2-cosθ)2. It doesn't look right. Also, I thought it might have been a little simpler to solve for sinθ and cosθ, then square, then add the equations and then implement TSny's hint.

amanda.ka said:
1.341cosφ = 2-cosθ
and
-1.341sinφ = sinθ

then square each term:
1.3412cos2φ = 22-cos2θ
and
-1.3412sin2φ = sin2θ

Note that (2 - cosθ)2 ≠ 22-cos2θ. In general (a + b)2 ≠ a2 + b2.

In order to use the trig identity cos2θ + sin2θ = 1, you could solve the first equation for cosθ. You already have an expression for sinθ, except I believe you have a sign error in -1.341sinφ = sinθ.

## 1. What is a 2D elastic collision?

A 2D elastic collision is a type of collision that occurs between two objects in a two-dimensional plane, where both the momentum and kinetic energy are conserved. This means that after the collision, the total momentum and total kinetic energy of the system remains the same.

## 2. How do you find the θ and φ angles in a 2D elastic collision?

The θ and φ angles can be found using the conservation of momentum and energy equations. These equations use the initial and final velocities of the objects, as well as their masses and angles of motion, to determine the values of θ and φ.

## 3. What are the steps to solve for the θ and φ angles in a 2D elastic collision?

The steps to solve for the θ and φ angles in a 2D elastic collision are as follows:

1. Identify and label the initial and final velocities of the objects.
2. Determine the masses and angles of motion of the objects.
3. Apply the conservation of momentum and energy equations.
4. Solve for the unknown angles (θ and φ).

## 4. Can the θ and φ angles be negative in a 2D elastic collision?

Yes, the θ and φ angles can be negative in a 2D elastic collision. This indicates that the objects' velocities are in the opposite direction of each other after the collision.

## 5. How is a 2D elastic collision different from a 2D inelastic collision?

In a 2D elastic collision, both the momentum and kinetic energy are conserved, whereas in a 2D inelastic collision, only the momentum is conserved. In an inelastic collision, some kinetic energy is converted into other forms, such as heat or sound, resulting in a decrease in the total kinetic energy of the system.

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