# Momentum: Derived Quantities

by Jared Rovny

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00:01 Let's take a quick example.

00:03 We have an object and it's going to be moving at a velocity V.

00:06 Suppose it hits a wall and it's going to have a force supplied to it by this wall because it's in collision with the wall.

00:12 We can suppose that collision lasts for a time delta T.

00:15 And then after it has collided with the wall, we assume that it bounces off and moves back away from the wall.

00:21 So in this situation, we can ask about what these derived quantities are just so we could see them sort of a plan an example.

00:28 So first we can define a coordinate system.

00:30 This is going to be specially important when we're considering vectors which momentum is.

00:35 So we have a positive X and a positive Y.

00:38 Our impulse is our change in momentum.

00:41 Initially, our object had a velocity V coming towards the wall.

00:45 And then finally, after it bounces off the wall, it has a velocity V away from the wall.

00:49 So we can write both of these momentum initially and finally.

00:53 We can just write the X component of our impulse because we only care about the X motion in this problem.

01:01 We do have to worry about the vectors though because initially the velocity is in the positive X direction and then the velocity is in the negative X direction after the collision.

01:10 So if we write the impulse, the final momentum minus the initial momentum, we can find the total impulse by saying the final momentum is minus mass times velocity where the minus comes from the fact that the velocity is in the negative X direction.

01:25 And then we have final minus initial, so then we do minus the initial momentum with the initial momentum, it's just M times V towards the wall in the positive direction.

01:35 So the first minus sign comes from the fact that the final momentum is a negative X direction, and the second minus sign in this impulse equation comes from the fact that we're subtracting.

01:46 We're doing final minus initial to find the change in momentum.

01:49 So for this reason, we have the impulse -- the total change in momentum is minus two, times the mass, times the velocity.

01:56 Since we found the impulse, so we found the change of momentum.

02:00 We can now ask what the force is on the object.

02:03 What force was exerted by the wall during this time delta T? We can find the average force of the entire period delta T by writing down the change of momentum over the changing time.

02:13 And since we've already calculated the change in momentum as minus two times the mass, times its velocity, we can just divide by the given quantity, delta T, and that would be our force.

02:23 For example, we could put in some numbers here.

02:26 Suppose we have a mass of 1 kilogram, a velocity of 2 meters per second, and the time of the collision of 4 seconds which should be a very, very long collision time -- look like a slow motion.

02:36 Then we would have an impulse of minus two times the mass, times the velocity, which should be minus 4 kilograms meters per second.

02:44 And we could also calculate the average force by plugging in the numbers to be: minus two MV divided by delta T, Where delta T again is considered to be 4 seconds in this sort of example here.

02:56 So the two and the two in the denominator, the numerator canceled with the four in the denominator, meaning that we have a force of --1 Newton.

03:04 So once again be careful with these signs.

03:06 We have a -1 Newton for our force because the force was acting in the negative X direction.

03:12 So force is a vector and it does have a direction, in this case, meaning the minus X direction.

### About the Lecture

The lecture Momentum: Derived Quantities by Jared Rovny is from the course Momentum.

### Included Quiz Questions

1. -7 kg∙m/s
2. +7 kg∙m/s
3. -3 kg∙m/s
4. +3 kg∙m/s
5. It depends on the time of the collision
1. 50 Newtons
2. 5 Newtons
3. Not enough information
4. 20 Newtons
5. 2 Newtons
1. 15 kg∙m/s
2. 10 kg∙m/s
3. 20 kg∙m/s
4. 5 kg∙m/s
5. -5 kg∙m/s

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