So let's do an example very quickly.
An object that is not just going to go straight into a wall and then bounce off
but is going to come in at an angle and bounce off of the wall.
So this object comes in and bounces off
leaving with the exact same speed and angle of incidence
when it bounces off that it came in with.
So in terms of the speed and the incident angle,
what was the impulse felt by the object?
And if the contact with the wall lasted some time delta T,
what was the force felt by the object?
So give this a shot and remember to keep momentum of vector.
And so if you are going to find the impulse or going to find the force,
you are going to need to remember to consider your vectors
and consider the different components of your vectors.
So let's see how this problem works.
And again we are going to keep in mind
that these are vector quantities.
So let's see what happens if we do this.
We have our object coming in.
We assume that it's coming in at some incident angle theta,
and it will be leaving at the same angle theta.
And we have the speed as V and magnitude and it has a direction,
so the velocity is the magnitude and the direction.
Since we are talking about vectors,
we should define a coordinate system.
Let's call this the positive X direction.
Let's called upwards the positive Y direction.
And now let's write our vectors.
So first let's write momentum as a vector.
And since it's hard to draw things in bold,
I'm just going to start using the arrow notation for vectors.
So this is momemtum as a vector is equal to.
Let's do the initial momentum first and let's write the vector of it's X and Y coordinates.
This object has a velocity that's going down which had an angle theta.
So if its velocity is downwards like this -- it's hitting a wall.
We have an angle theta here,
and it goes outwards like this,
then we can write the two components of this velocity vector as,
a vertical component downwards and a horizontal component to the right.
So this vector has two components.
One, its Y component directly downwards like this,
and one, its X component to the right like that.
So the X component of our momemntum vector is --
let's try it like this.
We have momentum equals mass times velocity as a vector initial,
so this is equal to mass.
And now we're going to write a velocity vector which we just saw.
If this hypotenuse is a length V,
then this part up here is opposite the angle theta.
And remember that we have sine of theta for things that are opposite the angle.
While this vertical component is downwards, so it's negative.
and it is the same in length as in this side,
which is the side touching the angle
-- it's a adjacent to the angle.
So this is V times the cosine of theta.
So we have our X component and our Y component -- let's write these down.
We have V times the sine of theta,
and now our Y component -- negative V times the cosine of theta.
So this our initial momentum vector.
Our final momemtum vector will be pretty well the exact the same with one difference, which is that,
notice this X direction got flipped.
It was pointing to the right up here,
and now it's pointing to the left in a negative direction.
So let's write out what these are.
We still have minus V cosine of theta for the vertical component of velocity.
The horizontal component of velocity has been reflected, we say,
so it's changed to minus V times the sine of theta in our coordinate system.
So now we have our new momentum vector.
P final is equal to mass times the final velocity,
which is equal to mass times minus V sine of theta, minus V cosine of theta.
So now we're ready to find the impulse.
Remember the impulse as a vector is simply final momentum minus the initial momentum.
So now we know exactly what this is.
We have final momentum is down here.
The initial momentum is up here.
So do each component of the vector bit by bit.
So let's first take the X component.
We have minus V sine of theta from the final minus from this negative sign here, the original sine of theta .
So now we have minus two V times the sine of theta.
So careful here and don't miss where all the negative signs came from.
P final has this negative sign,
and then we say minus P initial which itself is positive.
So we have a negative sign first from here and then a negative sign again from the subtraction step.
So if we put those two together, we have minus two, times V, times the sine of theta.
We do a comma and now we take the difference P final minus P initial of these guys,
but these are identical -- minus V cosine of theta, minus V cosine of theta.
So there's no change
If we do final minus initial, we're just subtracting the same thing from itself.
So now we have found our impulse vector.
So this is our now impulse.
It has two components -- an X component and a Y component.
So now if I wanted to write this an exam sitting by the way,
it's also possible to just write the X component and say,
the X component is minus two MV sine of theta.
And I can do calculations with just the X component,
and I could also write the Y component, the other component and say that that is equal to zero.
And this might be useful for calculations if you need to do any.
Second, we can say, "What is the force felt by the object?"
And we simply remember that our force equation is the change in momentum over the change in time.
When we have a change in time,
it's just delta T given to us over here on the side.
Since we've already calculated the change in momentum,
this is quite easy.
We just had one over change in time, times the change in momentum, M minus two V sine of theta and zero.
A simple way to write this might be like this.
You can put these constants M and delta T, write into your vector -- that's always okay.
So you have minus two M V over delta T, times the sine of theta, is the force and the X direction, and then nothing in the Y direction.
So using our vectors and our vector notation as well as this graph as given,
we can find both the impulse and the force felt by an object,
which is colliding with the wall even if it's colliding at some angle with that wall and then bouncing off.
Again, the only real tricky thing here was to remember our definitions,
but then also to remember to use vectors and to figure out what are vector components were
using the same trigonometry that we will have been using throughout this course and can continue to use.