# Impulse, momentum and confusion.

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Raeldor    254
I am trying to think out my collision response algorithm, but am a little confused about impulse and momentum. I was hoping someone could clarify my thinking. Firstly, let's say I have an object of 10kg that is stationary. I want to move it at 5km/h. To do this requires an external force. Am I correct in thinking that this external force is referred to as impulse? What form does this impulse take? Assuming the object has a velocity vector (which is currently zero), do I just add the vector with a length of 5 in the direction of the impulse to this? Should the impulse be added to the velocity as some kind of force*time delta? Assuming we have somehow applied this impulse to the object, does the then the velocity*mass become the new momentum of the object? So now we have an object with a velocity of 5 and a momentum of 50? Also, where does gravity come into play here? How is gravity applied to the object? Very confused. Thanks for any insight.

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etothex    728
impulse is defined as a force times a period of time, and causes a change in momentum.

The starting momentum of the object is zero, and you want it to have momentum of 50 kg*m/s (10kg * 5 m/s) Your change in momentum (delta-p) is 50 kg*m/s, so you need a force times time equal to that. It could be 5 N force over 10 seconds or 50 N over 1 second, both will cause the same change in momentum

gravity does not come into play, as long as the motion is always perpendicular to the gravitational field (you don't move up or down) and you ignore friction. If it is in the direction of gravity, you just add an acceleration of 9.8 m/s^2 downward, and using F=ma solve for the force vector down. Then you just need to make your impluse force the same as for without gravity but with the y-component larger by the force due to gravity(in the opposite direction)

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Guest Anonymous Poster
From an implementation standpoint, impulse means you change the momentum immediately (momentum += impulse). dt is assumed to be factored out as impulse is not continuous.

A continuous force, i.e gravity, is applied over time, e.g:

momentum += gravity * dt

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Raeldor    254
Thanks for the replys, I think I am slowly starting to understand. Some more questions though. Velocity is in m/s, yet gravity is 9.8 m/s^2. Why is the power of 2? Is this because the acceleration gets quicker and quicker, or is it because it is acceleration is meters per second per second? If it's the second can you just add 9.8 to velocity every second for example? Also, does the speed of an object under gravity just keep getting quicker and quicker until it reaches some kind of maximum because of air friction? I understand force and momentum are the same thing (ie, momentum is a force). What are these measured in? The formula is momentum=m*v, is this newtons?

Sorry for the bombardment of questions! Thanks for all the help!

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Zipster    2365
Acceleration is the rate of change of velocity, so you can think of it as meters per second, per second. You don't have to use 9.8, since that's in real-world units of meters. In a computer simulation it could be something else. And you're correct that an object keeps getting faster and faster until it reaches what it known as terminal velocity.

Finally, force and momentum are not the same thing. Force is mass times acceleration (Newtons), and momentum is mass times velocity (kilogram meters per second). If you multiply force by time, you'd get the units for momentum.

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Raeldor    254
Quote:
 Original post by ZipsterAcceleration is the rate of change of velocity, so you can think of it as meters per second, per second. You don't have to use 9.8, since that's in real-world units of meters. In a computer simulation it could be something else. And you're correct that an object keeps getting faster and faster until it reaches what it known as terminal velocity.Finally, force and momentum are not the same thing. Force is mass times acceleration (Newtons), and momentum is mass times velocity (kilogram meters per second). If you multiply force by time, you'd get the units for momentum.

Thank you. So is the normal order to add the acceleration to the velocity and then recalculate the momentum, or is there preferred way of doing it? Also, so 9.8 isn't the FORCE of gravity it's the ACCELERATION due to gravity?

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Zipster    2365
Usually you don't need to deal with momentum unless there's a collision, or some other transfer of energy. But you do want to add acceleration to velocity.

And you're correct, 9.8 is the acceleration due to gravity. If you recall Newton's law of universal gravitation, the gravitational force between two bodies is proportional to both their masses. However since force scales with mass, when you consider acceleration the mass of the second body (the one that isn't the Earth) it cancels out and you're left with a relatively constant acceleration.

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Raeldor    254
Quote:
 Original post by ZipsterUsually you don't need to deal with momentum unless there's a collision, or some other transfer of energy. But you do want to add acceleration to velocity.And you're correct, 9.8 is the acceleration due to gravity. If you recall Newton's law of universal gravitation, the gravitational force between two bodies is proportional to both their masses. However since force scales with mass, when you consider acceleration the mass of the second body (the one that isn't the Earth) it cancels out and you're left with a relatively constant acceleration.

Ok, that last paragraph went right over my head. I am not sure I understand the last sentence from a gramatical point of view.

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Zipster    2365
Yeah, that last sentence is definitely not a shining example of exemplary grammar, I apologize. [smile] It looked weird to me after I wrote it, but I had to run due to March Madness and all...

The gist of that paragraph was to take the law of universal gravitation, and replace force on the left side with mass times acceleration - you end up with ma = GMm/d^2. The 'm' cancels out on both sides and you're left with a = GM/d^2, which is independent of the mass of the body.

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Raeldor    254
Quote:
 Original post by ZipsterYeah, that last sentence is definitely not a shining example of exemplary grammar, I apologize. [smile] It looked weird to me after I wrote it, but I had to run due to March Madness and all...The gist of that paragraph was to take the law of universal gravitation, and replace force on the left side with mass times acceleration - you end up with ma = GMm/d^2. The 'm' cancels out on both sides and you're left with a = GM/d^2, which is independent of the mass of the body.

So you are saying here that this proves that the acceleration due to gravity is independent of the mass of the object being attracted (falling coconut for example) and is due entirely to the mass of the object it is being to (ie, the earth)?

So, to implement gravity then in my integration routine I need to calculate the change in velocity as

v=v+(-9.8f*dt)

where dt is the delta time, and then to update the position

x=x+(v*dt)

Does this sound right?