2D Elastic Collision Problem

Legit

1. The problem statement, all variables and given/known data

Let a particle of mass 2M have an initial velocity of v₀i (the i merely indicating it is traveling on the x axis) and undergo an elastic glancing collision with a particle of mass M initially at rest. After the collision, the M particle moves off at an angle of 45° above the + x-axis.

1. What is the speed of the M particle after the collision?

2. What is the velocity (magnitude and direction) of the mass 2M particle after the collision?

m₁ = 2M
m₂ = M
v_1i = v₀i
θ₂ = 45°

2. Relevant equations

p_ix = p_fx (momentum conservation in the x direction)
m₁v_1i = m₁v_1fcosθ₁ + m₂v_2fcosθ₂

p_iy = p_fy (momentum conservation in the y direction)
0 = m₁v_1fsinθ₁ + m₂v_2fsinθ₂

K_i = K_f (conservation of kinetic energy)
m₁v_1i² = m₁v_1f² + m₂v_2f²

3. The attempt at a solution

It appears to be a 3 equations 3 unknowns problem (the unknowns being v_2f, v_1f, and θ₁, which are the final velocities of the masses and the angle of mass 2M). I tried adding the two momentum equations to eliminate θ₁ and get a new equation to use with the energy equation, that way I would have two equations and two unknowns. But I ended up finding a value for v_1f to be -1.41v₀i, meaning the ball of larger mass collided with the ball of half-mass, then started going in the opposite direction with a larger velocity, which makes no sense... Any help would be appreciated.

ehild

The problem is wrong. The angle must be 30° or smaller.
If you solve the problem symbolically assuming an angle θ, you get a quadratic equation with the discriminant non-negative when cosθ≥√3/2.

ehild

Legit

So are you saying my professor wrote the problem wrong? That the problem is unsolvable if the angle is greater than 30°? I don't see why the angle must be 30° or lower....

Thanks for the reply by the way.

ehild

Isolate v_2fsin(θ₂) and v_2fcos(θ₂) from the equations for momentum and substitute into the energy equation. Try to solve for v₁ in terms of θ.

ehild

Legit

I don't understand. I need to isolate v_2fsinθ₂ and v_2fcosθ₂ in EACH momentum equation, then substitute into the energy equation? Am I isolating just v_2f or do I include the trig functions in my isolation? And how am I supposed to substitute? Do I substitute the result from isolating v_2fsinθ₂ and v_2fcosθ₂ separately, giving me two different energy equations?

ehild

Denote the components of the velocity of the smaller mass after the collision by v_2x and v_2y.
The momentum equations:

2Mv₀=2Mv_1f cos(θ)+Mv_2x
0=2Mv_1f sin(θ)+Mv_2y
The energy equation:
2Mv₀²=2Mv_1f²+M(v_2x²+v_2y²)

Isolating v_2x and v_2y from the first two equations:

v_2x=2v₀-2v_1fcos(θ)
v_2y=-2v_1fsin(θ).

Substituting into the energy equation:

2v₀²=2v_1f²+(2v₀-2v_1fcos(θ))²+(-2v_1fsin(θ))²

Expanding the squares and simplifying you arrive to the quadratic equation

6v_1f²-8v₀v_1fcos(θ)+2v₀²=0

whis has real solution for v_1f if cosθ≥√3/2.

Zeeshan86

What would be the case, if the ball is collided with the plate. Just like the table tennis.
Can we also consider the collision between the table tennis racket and ball as the 2D collision like collision between two ball ? ?
Will the shape of the plate will effect the equations ?

Zeeshan86

I am working on the 2D elastic collision problem.
In the attachment find my calculations.
First I solve the problem in forward direction and then I tried to solve the same problem with different angles.
The main problem is, I require another condition to calculate my last parameter.
Can anyone help me ...