(simple problem) Impulse force caused by cube collision?

[trogdor458]

I'm only interested in the 2d representation of this, as I will be applying whatever I learn to a physics engine

Anyways, I know that a block resting on a flat surface will have a normal force acting on it equal to its weight
However, I'm not interested in gravity

But, say a block comes flying in and strikes the ground
You know the velocity, angular velocity, and all the other measurable details about the collision
How might you determine the value of the impulse that the surface would apply to the block? (which I assume is pretty similar to the normal)

Also, am I correct to assume the impulse is perpendicular to the surface (as is with normal forces), and that total mechanical energy is conserved? (assuming no friction and completely elastic collision)

I'm in the middle of solving this problem, but am slow, so any help would be appreciated!

 

[kuruman]

Impulse is a vector given by $\vec J=\int \vec F~dt=\Delta \vec p$. Furthermore, the impulse imparted to the block is equal and opposite to the impulse imparted to the surface; Newton's 3rd law guarantees that. So for the vertical impulse imprted to the block, just find the change in the vertical component of the momentum. If the block is spinning before contact and not slipping on the surface while in contact, then angular impulse is also imparted on the block. It can be calculated because angular momentum about the point of contact is conserved since no torques about that point are acting on the bouncing object. Don't forget that angular momentum has two parts, angular momentum of the CM, $\vec L_{of~cm}=\vec r \times p$ and angular momentum about the CM, $\vec L_{about~cm}=I_{cm}\vec \omega$. Both parts are affected by the collision as some angular momentum about the center of mass is converted into angular momentum of the center of mass.
Example
If the object is spinning with initial angular velocity $\vec \omega_0$ and has CM velocity parallel to the surface, $V_x$, then its angular momentum just before the collision is $$L_{before}=I_{cm}\omega_0+mV_xd$$where $d$ is the distance from the point of contact to the CM. Because the contact is without sliding the object will start rolling on the surface meaning that its spin angular speed will increase/decrease from $\omega_0$ to $V'_x/d$ where $V_x'$ is the component of the velocity of the CM parallel to the surface after the collision. Then$$L_{after}=I_{cm}V'_x/d+mV'_xd.$$As you can see, in addition to the impulse perpendicular to the surface, there is impulse parallel to the surface. To find an expression for it (a) set $L_{before}=L_{after}$; (b) solve for $V_x'$ in terms of the other quantities and (c) use $J_{parallel}=m (V_x'-V_x)$.

If the spinning object slides while in contact, the situation becomes more complicated and I have already said enough.