Resultant velocities from 2D collision

In summary, the conversation discusses how to calculate the resultant velocities of two 2D circles colliding, assuming no elasticity or friction. The equations used are ρ=mv, m1v1 + m2v2 = m1v1` + m2v2`, and Speed = Square root of (X-velocity^2 + Y-velocity^2). It is also mentioned that the size of the circles is equal to their mass and that the collision is assumed to be elastic without any non-conservative forces. To calculate the resultant velocities, one must resolve the velocities along the center to center line of the spheres at the time of impact. The tangential component remains unchanged.
  • #1
Hurpadurp
2
0

Homework Statement


Suppose you have two 2D circles, where size ≈ mass, each of which is moving with known x- and y-velocities. The two collide, not necessarily head-on (one could broadside the other, etc). How do I calculate the resultant velocities, assuming no elasticity or friction?

Homework Equations


ρ = mv
m1v1 + m2v2 = m1v1` + m2v2`
Speed = Square root of (X-velocity^2 + Y-velocity^2)

The Attempt at a Solution


Each circle is moving in a particular angle (direction), and hits the other at a particular angle. If the difference between angles is 0, then the mower transfers all of its momentum to the other. If the difference is approaching 90 degrees, then the amount transferred is approaching 0.I apologize if this has already been answered; I couldn't find it.
Thanks!
 
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  • #2
Hurpadurp said:

Homework Statement


Suppose you have two 2D circles, where size ≈ mass, each of which is moving with known x- and y-velocities. The two collide, not necessarily head-on (one could broadside the other, etc). How do I calculate the resultant velocities, assuming no elasticity or friction?

Homework Equations


ρ = mv
m1v1 + m2v2 = m1v1` + m2v2`
Speed = Square root of (X-velocity^2 + Y-velocity^2)

The Attempt at a Solution


Each circle is moving in a particular angle (direction), and hits the other at a particular angle. If the difference between angles is 0, then the mower transfers all of its momentum to the other. If the difference is approaching 90 degrees, then the amount transferred is approaching 0.


I apologize if this has already been answered; I couldn't find it.
Thanks!

Hello Hurpadurp,
What do you mean by size ≈ mass?And what is the constraint "no elasticity"?Do you mean that the balls are identical and the collision is assumed to be elastic without any non conservative forces? If yes try to resolve the velocities along the center to center line of spheres at the time of impact.You will be done.
regards
Yukoel
 
  • #3
Yukoel,

Here's a little more information on why I'm asking this:
I'm making a program in which objects, represented by circles, are moving around on the screen. When they hit each other, a collision function is called - the size of the circle is the same as its mass; e.g. circles of radius 70 units have half the mass of circles of radius 140 units. For ease of use, let's say the units are meters, and convert nicely to kilograms; the speeds are in m/s.
I'm unfamiliar with the concept of elasticity (my recent college courses just dealt with head-on collisions with either negligible or a specific amount of friction). In here, there are no non-conservative forces.

I'm sorry, but what do you mean by resolving "the velocities along the center to center line of spheres at the time of impact?"

Thanks,
Hurpadurp

Edit: A little more information: I checked http://en.wikipedia.org/wiki/Elastic_collision#Two-_and_three-dimensional but I don't quite understand if it's explaining the situation for the second particle being at rest, or more generally.
 
Last edited:
  • #4
Hurpadurp said:
Yukoel,

Here's a little more information on why I'm asking this:
I'm making a program in which objects, represented by circles, are moving around on the screen. When they hit each other, a collision function is called - the size of the circle is the same as its mass; e.g. circles of radius 70 units have half the mass of circles of radius 140 units. For ease of use, let's say the units are meters, and convert nicely to kilograms; the speeds are in m/s.
I'm unfamiliar with the concept of elasticity (my recent college courses just dealt with head-on collisions with either negligible or a specific amount of friction). In here, there are no non-conservative forces.

I'm sorry, but what do you mean by resolving "the velocities along the center to center line of spheres at the time of impact?"

Thanks,
Hurpadurp

Edit: A little more information: I checked http://en.wikipedia.org/wiki/Elastic_collision#Two-_and_three-dimensional but I don't quite understand if it's explaining the situation for the second particle being at rest, or more generally.
Hello Hurpaderp,
Um I made this image here at http://img823.imageshack.us/img823/4985/f5461f6608dd48fdb74f6a2.png (I am bad at paint and photoshop so sorry :( )
I think it is kinda what your situation demands.The center to center line is the red line(in the picture).Resolve velocities along this line and apply expressions for final velocities on them like two masses colliding with given initial velocities.The tangential component(the component perpendicular to the red line) remains untouched.The wiki link shows a stationary equal mass being acted upon by the collision.This is why the final trajectories of both are perpendicular.
Hoping this helps.
regards
Yukoel
 
Last edited by a moderator:
  • #5


I would approach this problem by first breaking down the problem into its components. Since we are dealing with a 2D collision, we can use vector analysis to determine the resultant velocities.

First, we need to determine the initial velocities of each circle in terms of their x- and y-components. This can be done using trigonometry, where the x-velocity is equal to the speed multiplied by the cosine of the angle, and the y-velocity is equal to the speed multiplied by the sine of the angle.

Next, we need to determine the angle of collision between the two circles. This can be done by finding the angle between the line connecting the centers of the circles and the line perpendicular to the tangent of the point of collision.

Once we have the angle of collision, we can use the conservation of momentum equation (m1v1 + m2v2 = m1v1` + m2v2`) to determine the resultant velocities. We can do this by resolving the initial velocities into their x- and y-components, and then applying the conservation of momentum equation for both the x- and y-components separately.

Finally, we can use the Pythagorean theorem to find the magnitude of the resultant velocity, by taking the square root of the sum of the squares of the x- and y-components.

It is important to note that this solution assumes no elasticity or friction. In real-world scenarios, these factors may need to be taken into account, which would require additional equations and calculations.

I hope this helps in understanding how to calculate the resultant velocities from a 2D collision.
 

What is a resultant velocity?

A resultant velocity is the vector sum of all the individual velocities in a system. It represents the overall velocity and direction of an object after a collision or interaction with another object.

How is the resultant velocity calculated?

The resultant velocity can be calculated using the Pythagorean theorem, where the square of the resultant velocity is equal to the sum of the squares of the individual velocities. The direction of the resultant velocity can be found using trigonometric functions.

What is a 2D collision?

A 2D collision is a collision between two objects that occur in a two-dimensional plane, where the objects have velocities in two different directions.

What factors can affect the resultant velocities in a 2D collision?

The mass and velocity of the objects involved, as well as the angle and direction of their initial velocities, can all affect the resultant velocities in a 2D collision. Other factors such as friction and air resistance may also play a role.

What is the conservation of momentum in a 2D collision?

The conservation of momentum states that the total momentum of a closed system (where no external forces are acting) remains constant before and after a collision. In a 2D collision, this means that the sum of the momenta in the x-direction and the y-direction before the collision is equal to the sum of the momenta in the x-direction and the y-direction after the collision.

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