# Homework Help: Momentum graph given, find the launch angle, level ground

1. Jul 4, 2010

### Beamsbox

Calculus based course, this is from the "linear momentum of a system of particles" section.

I've attached a rough sketch of the graph from my textbook. The main points of the graph are this, it's a parabola, at T=0 and T = 4.5, P(momentum) = 6. At T= 2.25 (maybe) it bottoms out. I say maybe because it doesn't really say where exactly it bottoms out, just shows a graph... and it appears to bottom out between 2 and 2.5.

The question:

At Time t = 0, a ball is struck at ground level and sent over level ground. (The graph) gives the magnitude (p) of the ball's momentum versus time (t) during the flight. At what angle is the ball launched?

Possibly useful equations (just not sure which to use):
Pi = Pf = mivi = mfvf(conservation of momentum)

Fnet = dp/dt

P = Mvcom (com meaning the com)

My thoughts:
Not sure where to start, other than taking the derivative of the momentum, but I'm not sure how to find the equation of the parabola... I've come close, but even if I had it, thus converting it to the force, I'm not sure where the angle would come in... though I thinkif I could get to that point I could figure it out.

Thanks prior, all.

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2. Jul 4, 2010

### pgardn

Ahhhh you are the person that gets the tricky questions that are sort of misleading (ie they give you data you dont need). They are easier after closer inspection.

Ok So we agree that the momentum is large, then goes to a small number, and then large again. So lets just look at this as a ball in 2-d motion, projectile motion. I thing we can assume the mass of the ball is not changing but the velocity is. What do you know about the velocity of a ball "shot" at an angle at the beginning of its flight starting presumably and the ground, and at its max height off the ground? Vx and Vy and Vo... I think this is the way to start.

3. Jul 4, 2010

### Beamsbox

That's great... that you recognize me as the tricky question guy, I wouldn't be here if they weren't so... I'm always the lucky one... and yes, they are always a bit easier after we figure them out. ;) So I must once again thank you for your help.

Well, we know that the acceleration is a constant 9.8 m/s2 downward. Concerning Vy... so since we agree that the velocity is decreasing (as momentum decreases), and we know that at the top of the trajectory, Vo becomes 0 for but an instant. That means that at the bottom of the momentum curve, the object is at it's highest point. Say around 2.25 seconds into it.

But analyzing the components of their motions might help. (Like you said...)

Horizontally,
Xf - Xi = V0t
Vo = V0cos(theta)

so, Xf - Xi = V0cos(theta)t

Vertically,
yf - yi = V0sin(theta0)t-1/2gt2

Assuming (0) initial velocity, the ball shoots up 24.8 m, before it falls... but I don't see the usefulness of that...

Last edited: Jul 4, 2010
4. Jul 4, 2010

### pgardn

Well you know you have 6 units of momentum at "launch".

How many units of momentum do you have when the object is at its highest point, its not 6. And you said at one moment during the flight Vo was zero? If it was zero then you would have no momentum. Your graph says there is momentum thus velocity during the entire flight. But there is a moment during the flight when the velocity, thus the momentum, is the smallest. And the number is on your graph for the momentum at this point.

Oh and I did not mean to imply you make up these problems. Your "teach" is a bit tricky imo. Its still a question that I slapped myself once I saw it.

Last edited: Jul 4, 2010
5. Jul 4, 2010

### Beamsbox

Haha, no offense taken. And it's not the teach, they're straight from the book, they just happen to be the few I can't figure out. I guess I'm more of a straight-forward visualizer, ;).

Ya, I guess I was wrong when I said there was no momentum. But I assume that the momentum is only in the horizontal component at that point. Wouldn't that be right, you have no vertical velocity, just horizontal. Perhaps I can use that to my advantage somehow?

So, you have 4 units at the highest point, all directed horizontally. This would remain constant throughout the flight, as there are no external forces acting on the system. And since you have 6 units to begin with, and 4 of that are horizontal, then the vertical component must be the difference, which is 2.

So perhaps this is a stretch, but maybe it works.

Px0 = 4
Py0 = 2

Tan(Theta) = .5
Theta = Tan-1(.5)... which is not the answer I'm looking for.

Whoops, there it goes... my brain just shut-off on me.

6. Jul 4, 2010

### pgardn

So the velocity when the momentum is worth 4 units, is all Vx. So the total momentum at launch is 6. Momentum is m*v... So initial momentum, which is 6 is m*vo. How can you describe the smallest momentum, 4 units, in terms of m and vo... ? and you know that mass, m, stays the same throughout the flight. Its not a full vo of velocity when the ball has 4 units of momentum, yes?

when you stated calculus based, and they give you this graph, I took it to mean we would have to do some math, but there is no calculus involved in this particlar problem...

Last edited: Jul 4, 2010