Calculating impulse of a collision

[alterecho]

I'm trying to apply the equation from Wikipedia (http://en.wikipedia.org/wiki/Collision_response) to find the impulse of a collision. How do i find/calculate the surface normal n, which is a unit vector?

 

[mfb]

It depends on the implementation of your objects. Find the point of collision, and the part of the object which describes this, calculate the vector normal to the surface.

 

[alterecho]

How do i calculate the vector normal?

 

[mfb]

How do you know how your objects looks like?
Two vectors are orthogonal if their scalar product is 0.

Your question is similar to "how do I calculate the speed of a car". You have to say what you know about the car (for example: "traveled x km in y minutes").

 

[alterecho]

Here's an example of the situation I'm working with.

The distance vector for point of intersection would be (0, -20) in this case for the ball.
I assume the ball as the 1st object and the ground as the second object. I'm also considering the mass of the ground to be infinite and mass of the ball to be 1 and its moment of inertia as 400.

 

[mfb]

The ground has a horizontal surface in the x-direction and you have a two-dimensional problem? In this case, the orthogonal vector goes in the y-direction. It is simply (0,1).

 

[alterecho]

I did use that. Suppose i have,
$V_{i}$ = (20, -10),
$Vp_{i}$ = (20, -10),
$ω_{i}$ = 0,
n = (0, 1),
$M_{a}$ = 1,
$I_{a}$ = 400,
$r_{a}$ = (0, -20),
I get the $V_{f}$ as (20, -30).

This is the procedure i follow,
-Calculate the impulse j,
-Apply the formula $V_{f}$ = $V_{i}$ - $\frac{j}{M}$ * n (as given in Wikipedia).

The result is obviously wrong since with this final velocity, the ball will fall through the floor.
Am i missing some step?

 

[mfb]

I think you have a sign error for j. Using your sign convention, it should be negative.

 

[alterecho]

You mean $V_{f}$ = $V_{i}$ - $\frac{-j}{M}$ * n ?
But we are taking into consideration the normal of the impulse so how is that?

 

[haruspex]

In the wikipedia diagram, the normal vector points away from object 1. This leads to the minus sign in the web page equation. You should have V_f = V_i + (positive factor)n.

 

[alterecho]

You're right. But when i apply the initial values as:
e = 1
$V_{i}$ = (10, -20)
$W_{i}$ = 0
r = (0, -20)
$V_{p}$ = (10, -20)
n = (-1, 1)
m = 1
I = 200

For Collision with a Wall/ground, I'm using equation from this site (modified version of the Wikipedia equation): http://www.myphysicslab.com/collision.html

j = $\frac{−(1 + e) V_{p1} · n}{\frac{1}{m} + \frac{(r × n)^{2}}{I}}$

I get
j = 59.4,
$V_{f}$ = (-49.4, 39.4)
$W_{f}$ = -5.9

This obviously seems wrong. What should i do?

 

[haruspex]

You're right. But when i apply the initial values as:
e = 1
$V_{i}$ = (10, -20)
$W_{i}$ = 0
r = (0, -20)
$V_{p}$ = (10, -20)
n = (-1, 1)
m = 1
I = 200

For Collision with a Wall/ground, I'm using equation from this site (modified version of the Wikipedia equation): http://www.myphysicslab.com/collision.html

j = $\frac{−(1 + e) V_{p1} · n}{\frac{1}{m} + \frac{(r × n)^{2}}{I}}$

I get j = 59.4,

|r x n| = 20, agreed? This gives me j = (2*30)/(1/1+400/200) = 20.

 

[alterecho]

Sorry for the late reply. I just identified an error in my application of the equation. I was taking the square AFTER dividing (r x n) by I in the denominator.

I'll have to see how this turns out in my application. Will do some tests.

You say |r x n| = 20. Do we need to take the absolute of the result of the cross product?