Physics: momentum: bouncing off the wall

[pshla619]

---

Members don't see this ad.

Hi guys,

I have a quick question on a momentum problem.

"An atom whose momentum has magnitude p and where velocity vector is perpendicular to an infinitely heavy wall, strikes the wall and recoils elastically. What is the magnitude of the change in the atom's momentum?"

Here is my work:
MatomVatom (before) + MwallVwall (before) = MatomVatom (after) + MwallVwall (after)

Both wall momentums are zero because wall does not move..
So mathematically, Velocity of the atom before and after are the same?
But conceptually, velocity of the atom after it hits the wall should have a negative sign.
What am I missing here?

Answer key just said that since change in P = Pafter - Pbefore = (-P) - (P) = -2P.
And since the question asked for magnitude, the answer was "2P."

I get it conceptually, but mathematically, I can't see how to make the after velocity of the atom be negative..
Thanks guys!

 

[Deadlifts]

Hi guys,

I have a quick question on a momentum problem.

"An atom whose momentum has magnitude p and where velocity vector is perpendicular to an infinitely heavy wall, strikes the wall and recoils elastically. What is the magnitude of the change in the atom's momentum?"

Here is my work:
MatomVatom (before) + MwallVwall (before) = MatomVatom (after) + MwallVwall (after)

Both wall momentums are zero because wall does not move..
So mathematically, Velocity of the atom before and after are the same?
But conceptually, velocity of the atom after it hits the wall should have a negative sign.
What am I missing here?

Answer key just said that since change in P = Pafter - Pbefore = (-P) - (P) = -2P.
And since the question asked for magnitude, the answer was "2P."

I get it conceptually, but mathematically, I can't see how to make the after velocity of the atom be negative..
Thanks guys!

Velocity is a vector quantity. It has a direction in addition to its magnitude.
Define +x as towards the wall.
initial
(atom)--\--> |Wall|

final

<--\--(atom) |Wall|

The final velocity is negative because it is in the opposite direction of the initial velocity.

 

[pshla619]

Velocity is a vector quantity. It has a direction in addition to its magnitude.
Define +x as towards the wall.
initial
(atom)--\--> |Wall|

final

<--\--(atom) |Wall|

The final velocity is negative because it is in the opposite direction of the initial velocity.

Thanks for the reply. But yes, I know that velocity should be + initially/before and - after.
But looking at the change in total momentum equation, they should be the same mathematically.

MballVball + MwallVwall = MballVball + MwallVwall
left side = before. right side = after
Velocity of wall is zero so both of them are zero.
That leaves us with MballVball = MballVball
So mathematically, (obviously I'm missing something) velocity are the same before and after.

:/

 

[milski]

Can't type it on the phone but from conservation of momentum and conversation of energy you should be able to fine the velocity of the atom after impact as a function of the two masses and the initial speed of the atom. Take the limit of that for the mass of the wall approaching infinity and you should get that vafter=-vbefore.

 

[pshla619]

Can't type it on the phone but from conservation of momentum and conversation of energy you should be able to fine the velocity of the atom after impact as a function of the two masses and the initial speed of the atom. Take the limit of that for the mass of the wall approaching infinity and you should get that vafter=-vbefore.

Thank you for the reply. Could you explain a bit further? Because no matter what the mass of the wall is, since the wall is not moving, isn't before and after momentum of wall just zero?
Leaving before and after momentum of the atom/ball exactly the same? mathematically that is.. conceptually I know/understand that after velocity should be negative and before should be positive.

 

[milski]

Mathematically you're trying to use that infinity minus infinity is zero and that's not correct. The trick is to find an expression depending on the mass which does have a limit when the mass approaches infinity. If you cannot do the equations from my previous post, I'll post details tomorrow.

For what you are trying to do, you cannot say that the speed of the wall does not change. The change may be infinitesimal but mathematically that's not the same thing as zero.

 

[pshla619]

I'll post details tomorrow.

For what you are trying to do, you cannot say that the speed of the wall does not change. The change may be infinitesimal but mathematically that's not the same thing as zero.

Ah, ok. So my error was that I assumed velocity of wall to be zero.
I'd appreciate details of work if you can post tomorrow or whenever you get a chance! 🙂

 

[milski]

Ah, ok. So my error was that I assumed velocity of wall to be zero.
I'd appreciate details of work if you can post tomorrow or whenever you get a chance! 🙂

Right, it is zero for all practical purposes but it is not really zero for mathematical purposes. Tribulations like yours lead lead to the invention of calculus. 🙂

I'll post something tomorrow - I have only a phone with me now and no scratch paper.

 

[milski]

 

[adrakdavra]

Impulse is (change in M.V = F . change in Time)
For example if the mass = 1 kg and V = 10m/s , if the mass hits the wall and then stops right in that specific time without bouncing back then the change in P is zero, however because the mass is flying back after bouncing off the wall then we are gaining momentum , for example if the same 1 kg mass bounce back at 10 m/s then our change in momentum = final - initial = 1 kg . -10m/s - ( 1kg . +10m/s) = - 20, the negative is because direction changed

 

[pshla619]

Thanks a lot for the reply! I really appreciate it