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Momentum an dimpulse

  • Thread starter ACLerok
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  • #1
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Right now we're studying momentum and impulse. I know how to find momentum and impulse but how do you find the total magnitude of momentum in a system? How about the change in momentum?

And finally, for example, if two objects of different mass and different speed collide with each other and then stick together, how do you find their final velocity and in what direction?
 

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  • #2
Doc Al
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Originally posted by ACLerok
Right now we're studying momentum and impulse. I know how to find momentum and impulse but how do you find the total magnitude of momentum in a system? How about the change in momentum?
To find the total momentum of a system, add the momenta of each object. Remember that momentum is a vector, so learn to add vectors.
Also, Δmomentum = momentumfinal-momentuminitial. For this, you have to subtract vectors.
And finally, for example, if two objects of different mass and different speed collide with each other and then stick together, how do you find their final velocity and in what direction?
You take advantage of the fact that momentum doesn't change during a collision. So the initial momentum (vector sum of the two momenta before collision) must equal the final momentum (massXvelocity of the combined mass after the collision).

But, talk is cheap. The only way to learn this stuff is by doing a bunch of problems. Get busy!
 
  • #3
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To add vectors you square both components (or all three if z is involved) and take the square root of the some of them correct?
Thanks!
 
  • #4
Doc Al
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Originally posted by ACLerok
To add vectors you square both components (or all three if z is involved) and take the square root of the some of them correct?
There are at least two ways to add vectors. You can break them into components, and add the components of each. The magnitude of the resultant is found as you state:

R = √(Rx2 + Ry2)
But you also have to find the direction of the resultant vector.

The other way is to draw the arrows and add them directly, using trig. Sometimes that's easier.
 
  • #5
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Originally posted by Doc Al
There are at least two ways to add vectors. You can break them into components, and add the components of each. The magnitude of the resultant is found as you state:

R = √(Rx2 + Ry2)
But you also have to find the direction of the resultant vector.

The other way is to draw the arrows and add them directly, using trig. Sometimes that's easier.
k i understand.. the direction and angle can be found just by finding the arctan of the y component divided by the x component. thanks
 
  • #6
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Me again. So I was able to do many of the word problems but there is one about momentum and impulse that is till buggin the crap out of me. Basically, a ball is dropped from a certain height, its a table, and then bounces back up. The problem is asking to find the momentum before and after the collision with the table. Momentum is massXvelocity but theres is no mention of velocity. I tried using the equation for gravitational potential energy (mgh) but that is wrong. Any help please? Once I find the momentum before, howdo I find it after the collision???
Thanks
 
  • #7
Doc Al
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There's nothing wrong with using "mgh" to find the speed before the collision. KE = mgh. Once you have the KE, calculate the speed from the definition of KE. Given the speed, what's the velocity and thus the momentum?

As far as after the collision, that depends on the kind of collision. Was it elastic? (Did it bounce back up to the original height?) If so, energy is conserved. If it has the same speed just after the collision, what about its velocity and momentum?

Remember that momentum and velocity are vectors.
 
  • #8
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it does not bounce back to its original drop point. momentum is not conserved. how would i tackle this?
 
  • #9
Doc Al
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Originally posted by ACLerok
it does not bounce back to its original drop point. momentum is not conserved. how would i tackle this?
How high does it bounce? If you know that, you can figure out how fast it left the ground. For example, if it doesn't bounce at all, all the KE is lost: its speed after the collision is zero. (Think of dropping a lump of putty.)

Momentum conservation has nothing to do with it; we're talking about energy conservation.
 
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