Elastic Collision in two dimensions question

In summary: That leaves you with two equations that are identical except for the unknown variable, θ. You can use the Pythagorean theorem to solve for θ.In summary, the first ball moves away with angle 30° to the original path, so its speed is 3.0 m/s after the collision.
  • #1
iwatchcredits
3
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Homework Statement

[/B]A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown (The first ball is traveling to the right). If the first ball moves away with angle 30° to the original path, determine
  1. the speed of the first ball after the collision.
  2. the speed and direction of the second ball after the collision.

Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

The Attempt at a Solution


[/B]
So far I understand that the conservation of momentum and energy can be used, however with the amount of different ways these formulas can be used I don't know where to go with them. Does anyone have any advice to help break this down algebraically?
 
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  • #2
iwatchcredits said:

Homework Statement

[/B]A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown (The first ball is traveling to the right). If the first ball moves away with angle 30° to the original path, determine
  1. the speed of the first ball after the collision.
  2. the speed and direction of the second ball after the collision.

Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

The Attempt at a Solution


[/B]
So far I understand that the conservation of momentum and energy can be used, however with the amount of different ways these formulas can be used I don't know where to go with them. Does anyone have any advice to help break this down algebraically?
Remember that momentum is vector, so it is conserved in all directions. That should give you two equations, plus the energy equation.
Pick an xy system of coordinates. Either create separate variables for velocities in different directions (x and y) or introduce variables to represent angles. Then rewrite the relevant equations using these.
 
  • #3
The formulas I have come up with are

- v1'sin theta = v2'sin theta
v1 - v1'cos theta = v2'cos theta

and
v1^2 - V1'^2 = V2'^2

I am unsure of how I join these equations though, I know I need to square the momentum equation and sub them into the kinetic equation but I'm not quite sure how, do I need to divide over the cos' and sin's to get them on the same side or what? (v2' = -v1'sin theta/sin theta)
 
  • #4
iwatchcredits said:
- v1'sin theta = v2'sin theta
v1 - v1'cos theta = v2'cos theta
You are using the same angle, θ, for V1' and V2'. It was given in the problem statement that the angle of V1' is 30°.

Edit: Welcome to Physics Forums. Except for using the same angle θ for both V1' and V2', I think your equations are correct.
 
  • #5
Ok, I am still unsure of how to continue on this question though, I don't really even know what I'm trying to do next (get rid of the angles?) but I don't know how to join the equations properly and it seems no matter what algebra I do I just end up going in a circle.
 
  • #6
iwatchcredits said:
I don't really even know what I'm trying to do next (get rid of the angles?)
There is one unknown angle, θ, which is the angle of V2'. The V1' angle is 30°.

I remember solving this type of problem before, and yes, it is kind of a tricky solution. I am heading out the door in about 5 minutes so don't have time right now. But it definitely involves one or two trig. identities. The first one may involve sin2 + cos2 = 1, but I don't remember for sure.
 
  • #7
iwatchcredits said:
Ok, I am still unsure of how to continue on this question though, I don't really even know what I'm trying to do next (get rid of the angles?) but I don't know how to join the equations properly and it seems no matter what algebra I do I just end up going in a circle.
Take this into account:
iwatchcredits said:
an identical stationary ball
That simplifies the equations hugely.
There is a standard procedure for solving simultaneous equations.
Take stock of which variables are known and pick one other which is to be determined. You are asked first for v1' so let's pick that. Now pick one of the remaining variables, call it x, as a target for elimination. Rearrange one of the equations in the form x=... and use that to get rid of x in the others.
 

FAQ: Elastic Collision in two dimensions question

What is an elastic collision in two dimensions?

An elastic collision in two dimensions is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy of the objects before the collision is equal to the total kinetic energy after the collision.

What are the conditions for an elastic collision in two dimensions to occur?

The conditions for an elastic collision in two dimensions to occur are that there is no external force acting on the objects, the objects are not permanently deformed, and there is no loss of kinetic energy.

How is momentum conserved in an elastic collision in two dimensions?

In an elastic collision in two dimensions, momentum is conserved because the total momentum of the objects before the collision is equal to the total momentum after the collision. This means that the vector sum of the momentums of the objects in the x and y directions remains constant.

What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, there is no loss of kinetic energy, while in an inelastic collision, there is some loss of kinetic energy. In an inelastic collision, the objects may also become permanently deformed.

How can the coefficient of restitution be used to determine if a collision is elastic or inelastic?

The coefficient of restitution is a measure of the elasticity of a collision. If the coefficient of restitution is equal to 1, the collision is elastic. If the coefficient of restitution is less than 1, the collision is inelastic.

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