Change in momentum when a body is thrown up and falls back down.

[s0ft]

Say, a body of mass 'm' is thrown at a certain angle with the vertical with certain initial velocity 'u'.
The initial momentum of this object is mu.
Am I right to say that if it spent 't' seconds in flight and came back, its final velocity will still be 'u'?
And the change in momentum will be mu-(-mu)=2mu?
And would this be correct or not:
F=dp/dt
=>dp=F*dt
Since F=mg and dt=t, change in momentum=mgt?
One of these has to be wrong. I don't know which one.

 

[Doc Al]

Say, a body of mass 'm' is thrown at a certain angle with the vertical with certain initial velocity 'u'.
The initial momentum of this object is mu.

OK.

Am I right to say that if it spent 't' seconds in flight and came back, its final velocity will still be 'u'?
And the change in momentum will be mu-(-mu)=2mu?

No, because only the vertical component of the velocity will be reversed. The speed will be the same, but the final velocity will not simply be the negative of the initial velocity. (Also not that change is always final minus initial.)

You can find the change in momentum, but you'd have to treat the velocity as a vector.

And would this be correct or not:
F=dp/dt
=>dp=F*dt
Since F=mg and dt=t, change in momentum=mgt?

That's correct.

One of these has to be wrong. I don't know which one.

The first was wrong because you confused speed with velocity.

If the body had been thrown straight up, instead of at an angle, then both methods would give the same answer.

 

[s0ft]

Got it! So treating the (change in momentum) as vector, this should be right:
Δ\vec{p} = \sqrt{(Δp_x)^2 + (Δp_y)^2}<br /> = \sqrt{(musinθ-musinθ)^2 + (mucosθ-(-mucosθ))^2}<br /> = 2mucosθ<br /> = mgt
I'm not sure if this is the most methodical way but how would one do it purely vectorically, rigorously?
Would it also be possible to do this taking the difference between vectors \vec{p_2} and \vec{p_1}? In that case what would be the angle between them?
And if the momentum was taken as scalar, then the change would be 0 right?

 

[Doc Al]

Got it! So treating the (change in momentum) as vector, this should be right:
Δ\vec{p} = \sqrt{(Δp_x)^2 + (Δp_y)^2}<br /> = \sqrt{(musinθ-musinθ)^2 + (mucosθ-(-mucosθ))^2}<br /> = 2mucosθ<br /> = mgt

Good.

I'm not sure if this is the most methodical way but how would one do it purely vectorically, rigorously?

What you did was a perfectly OK way to subtract vectors.

Would it also be possible to do this taking the difference between vectors \vec{p_2} and \vec{p_1}?

Isn't that what you just did?

In that case what would be the angle between them?

The same as the angle between the velocity vectors. (Figure it out by drawing a diagram for yourself.)

And if the momentum was taken as scalar, then the change would be 0 right?

Yes, the magnitude of the momentum is the same when it falls back down to the original height.

 

[s0ft]

Isn't that what you just did?

What I did, what I view it as, was treat the change in momentum as a vector and obtain its magnitude by using the change in X-component and change in Y-component, just like I'd do to get the magnitude for any vector whose X and Y components I know.

Would it also be possible to do this taking the difference between vectors p2→ and p1→?

Here, I mean by using the general vector expression for resultant of two vectors(in this particular case, -p1 and p2) a certain angle separating the two.
And though I know the result I will obtain should be exactly the same as that I got from the previous method, I just don't get what the angle between p1 and p2 is.

 

[Doc Al]

And though I know the result I will obtain should be exactly the same as that I got from the previous method, I just don't get what the angle between p1 and p2 is.

Draw yourself a diagram.

Hint: Initially, the velocity vector makes some angle above the horizontal. When it returns, the velocity vector makes the same angle below the horizontal.

 

[s0ft]

That works. Thanks a lot.
But I wish I could be as confident as you in saying your second statement. I had thought if that was the case but I simply couldn't take it as granted.
It would be nice to know how you were able to say for sure that it is so.

 

[Doc Al]

But I wish I could be as confident as you in saying your second statement. I had thought if that was the case but I simply couldn't take it as granted.
It would be nice to know how you were able to say for sure that it is so.

Draw a diagram. You know that the vertical component of velocity reverses, but the horizontal remains the same.

 

[s0ft]

Alright, assuming the x-component of the initial velocity to be unchanged and y-component being exactly reversed in direction but equal in magnitude, I could find the angles match.
Thank you.