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Physics: Conservation of momentum, 2d Collisions....HELP ME!!

C

chiwawa626

Lifer
I got this lab on conservation of momentum, 2d collisions one ball travels down a ramp and hits another at the end of the ramp (the ramp is assumed frictionless) and both balls fly off the table after collision. i need to show conservation of momentum:
m1v1 + m2v2 = m1v'1 + m2v'2
I have masses, displacment, initial velocities angles of balls. So i need to find the velocities after the collision or atleast one then i can get the other. BUT HOW? hmm wouldnt both balls have the same vertical velocity? since they fall about the same hieght?
 
Well, you know their vertical velocity right? So you know how long it took them to get to the floor. So based on how long that took and how far they traveled, you can determine the horizontal velocity. To be honest, I forget the exact equations, cuz we did this back in like October....are you guys seriously just doing it now?
 
hmm i think i got it..but wait..howmuch percent error do u think i should have since its done on such a small scale...im seeing my before total momentum is like .40 and after is like .25
 
man please we need a physics genius in here....i gota do these hard ass calculations for 18 trials..just wanna be sure their rite first..
 
You need to find out the following two things:

1. Did you know the distance from the end of the ramp to the floor?
2. Did you know where one of the balls landed on the floor?
 
just do this:
calculate the velocity of the ball on the ramp with this formula:
Vf^2 = Vi^2 + 2ad (a is 9.8 or 9.81 and d is the vertical height of the ramp)
now that we have the velocity of the first ball lets see its momentum by doing this:
p = mv which is the TOTAL MOMENTUM IN THE REACTION

you said BOTH BALLS fly off that table which means they had to give you 1 final velocity (otherwise there is no way to figure out how elastic the collision was) so do this:
p = mv to figure out the momentum of one of the balls assuming you know the final velocity on one of them while they fly off the table
now subtrace that second p from the first p (total momentum - momentum of 1 ball) which will give you the momentum of the second ball, now do this:
v = p/m because you said you knew the mass of both balls you know what m is and that p is the remaining p after you subtraced that other p from the total momentum and you just solve from there

i love physics 🙂
 
You can determine the velocity of the first ball right before collision by using conservation of energy. The ball has a potential energy at the top of the ramp (p1) and a kinetic energy at the bottom of the ramp (p2). If you use the table height as the reference, then all the potential becomes kinetic, so p1 = .5*m*(v^2), and you can find the velocity of the first ball before collision, which gives you the initial momentum. Since you know the time of fall, (or can figure it out from the height of the table where h=.5*g*t^2), and you know the horizontal distance covered by the ball(s), you can figure out their initial velocities (v=d/t).
 
Use:
y(t) = 0.5at^2 + vt + y, y(t) = final height, which is 0 in this case; a = -9.81 m/s or -32.2 ft/s, v = initial vertical velocity, which is 0 in this case when the ball
leaves the ramp; y = initial height from floor. Your unknown is t.

Then, the above equation reduced to: y = -0.5at^2 or t = sqrt(-2ay).

After you've found t, plug it in the following equation,

x(t) = vt, x(t) is the horizontal distance between the point on the floor right below the end of the ramp and where the particular ball lands.
Solve for v that is either v'1 or v'2, depends on which ball you've chosen to do the analysis on. Good luck. 🙂
 
jeez, in the text these messages are posted in, goodoptic's formula looks like how to get sine or cosine or something even though it's a simple formula lol (simple if you know it)
 


<< just do this:
calculate the velocity of the ball on the ramp with this formula:
Vf^2 = Vi^2 + 2ad (a is 9.8 or 9.81 and d is the vertical height of the ramp)
now that we have the velocity of the first ball lets see its momentum by doing this:
p = mv which is the TOTAL MOMENTUM IN THE REACTION >>



You cannot calculate the velocity of the ball this way. The acceleration (down the ramp) is not equal to g here, it is equal to g*sin(angle) of the ramp because there is a vertical and horizontal component to the weight of the ball.

You can find the final velocity using a=g*sin(angle) and the distance down the ramp, but conservation of energy is simpler.
 


<<

<< just do this:
calculate the velocity of the ball on the ramp with this formula:
Vf^2 = Vi^2 + 2ad (a is 9.8 or 9.81 and d is the vertical height of the ramp)
now that we have the velocity of the first ball lets see its momentum by doing this:
p = mv which is the TOTAL MOMENTUM IN THE REACTION >>



You cannot calculate the velocity of the ball this way. The acceleration (down the ramp) is not equal to g here, it is equal to g*sin(angle) of the ramp because there is a vertical and horizontal component to the weight of the ball. >>


actually you can, if the surface is frictionless then you can just use the straight height without using height as a ratio relating to sine

if you look at it, if you took a ramp that was 10m high but 20m long (low angle) and 10m high and 10m long. because the height is the same they will both give the exact same velocity but the low angel one would give a much lower acceleration on the ball itself but for a longer period of time. it's like saying i accelerate my car at 5m/s^2 for 2 seconds or 10m/s^2 for 1 second. greater acceleration but the same velocity in the end.
it's because the sine will just factor itself out anyway 😀
 
You're talking about motion in one-dimension. Because the ramp is at an angle, the weight of the ball is split into vertical and horizontal components. As the angle decreases, more of the weight of the ball in the vertical direction is countered by the normal force of the ramp. The velocity of the ball also has two components. The component along the ramp is what we need to find, since the vertical component will be stopped by the table at the end of the ramp and is independent of the horizontal component.

The total velocity (v^2 = horizontal^2 + vertical^2) of the ball is the same as if it were dropped from the same height - that's where the sines and cosines factor out of the equation (sin^2 + cos^2 = 1).

EDIT: Of course, nobody here remembered to include rotational kinetic energy, but I'm guessing they were told to ignore that.
 
no dude you are taking friction into account man, the angle of the thing does not matter at all if there is no friction. im serious man
 
No, I'm not taking friction into account. If I were, you would have to multiply the normal force by a coefficient. I think you're thinking about the work done by the ball. It's equal (on a frictionless surface) to if the ball were dropped, because gravity is a conservative force. That's why the total velocity is the same (v^2 = horizontal^2 + vertical^2), but here we need the component along the ramp, which is less than the total velocity as you can see by the equation in this sentence.
 
interesting theory or fact or whatever... the name is not the point
it sounds air tight but im tired and i have a question. i drew a diagram here and figured out some things:
if the ramp was 10 high and 5 wide and the ball was 5kg. the force straight down would be 49N and the angle of the ramp would be 63.43
using some trig i found that the force of gravity straight down would be 49N (5kg x 9.81m/s^2 = 49N)Fparallel is 54.79N due to the force of gravity (cos26.57 = 49/x turns into x = 49/cos26.57)
my testing shows that what you are saying is correct in terms of force in direction. i lost my train of thought though so i don't know how to turn these into velocities.... too tired

i think i made an error there because it makes no sense how the indirect force of gravity after it's direction is changed is GREATER than the direct force of gravity going straight down
if you have figured out WHERE i went wrong plz tell me. im gonna lose sleep over this

 
The force down the ramp should be 43.8N (5kg*9.8m/s^2*sin(63.44)) and the force perpendicular to the ramp should be 21.91N (5kg * 9.8m/s^2*cos(63.44). The force of the block straight down is then the square root of (43.8N)^2 + (21.91N)^2 which is 49N (5kg*9.8 m/s^2). The angle (theta) of the ramp is the same as the angle between the weight vector (m*g) pointing straight down and the component of the weight pointing down perpendicular to the ramp(directly opposite to the normal force). This can be shown through similar triangles but I don't really know how to do that by typing.
 
Another question:....suppose the ball that was stationary that was beign hit was at an angle...that wouldnt make a diffrence...but why ?
 
Andi had one question inthe lab i dont know what to put for: Why can u use projectile motion to investigate coservation of momentum in 2 dimensions?


and one more thing, if things are falling and moving forward at diffrent angles, isnt that momentum in 3 dimensions?
 


<< Another question:....suppose the ball that was stationary that was beign hit was at an angle...that wouldnt make a diffrence...but why ? >>



One ball would move off in one direction with a velocity; the other would move off in a different direction with another velocity. Think about hitting a pool ball to cut it into a pocket. The cue ball will alter its path after the collision.

I'm not quite sure what you're asking on the other two questions - could you give an example?

Edit: On the projectile motion question, it's because the projectiles are subject to a net external force - namely, gravity. For conservation of momentum to hold, there cannot be a net external force on the system.
 


<< hey heisenburg, if you have a scanner could you plz scan a sketch of the problem and send it to me at chooco6@icqmail.com

plz man? 🙂 >>



I don't have a scanner at my house, but there's one at my school I can use. I'll probably be up there tommorow or the next day, if you don't mind waiting a little while.
 
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