More silly physics fun

[Patrick del Poker Grande]

Inspired by Daryn's post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&Number=4011785&page=0&vc=1) in this awesome thread (http://forumserver.twoplustwo.com/showflat.php?Cat=0&Board=exchange&Number=4011389&p age=0&fpart=all):

This is a popular question in freshman physics. Imagine this set-up (because I don't want to take the time to MS Paint it): You have a wall at the front end of a cart and you're standing on the back end of this cart facing the wall and you have a bunch of perfectly elastic tennis balls. You throw the balls against the wall and they bounce off. Describe the motion of the cart, assuming perfect bearings and a smooth, no-slip condition between the flat ground and the wheels.

If you can't figure out what I'm talking about, imagine Daryn's fan on skateboard picture and replace the fan with a little person throwing tennis balls against the "sail".

[pokerdirty]

don't make OOT think (http://forumserver.twoplustwo.com/postlist.php?Cat=&Board=scimathphil) during the holidays

[daryn]

and no more "hand waving"

draw diagrams!! give reasoning!

[kurosh]

After you throw the ball, the cart will roll backwards. When the ball hits the wall, the cart will roll forwards. If you catch the ball, the cart will stop.

Eh, I think this might be right.

[Patrick del Poker Grande]

[ QUOTE ]
don't make OOT think (http://forumserver.twoplustwo.com/postlist.php?Cat=&Board=scimathphil) during the holidays

[/ QUOTE ]
Bah!

haha nice...OOT is hopeless on this one.

[MrWookie47]

In white? Eh, I guess so.<font color="white">

In the case of tennis balls, they'll bounce off the wall and will end up traveling in the direction I'll call -x. Since the system started with no momentum, and now a portion of the system has momentum in the -x direction, the cart/you/wall unit must gain momentum in the +x direction.

In the case of the fan system in the picture, though, the air molecules will be traveling in the +x direction as the air flows around the wall. Thus, the cart will gain momentum in the -x direction. </font>

Edit: <font color="white">In the brief time between when you release the ball and when the ball hits the wall, you travel in the -x direction. </font>

[Shajen]

how large is the wall?

How big is the shopping cart?

How hard am I throwing the ball against the wall?

What is the air speed of an unladen swallow?

[Matt Flynn]

[ QUOTE ]
Inspired by Daryn's post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&amp;Number=4011785&amp;page=0&amp;vc=1) in this awesome thread (http://forumserver.twoplustwo.com/showflat.php?Cat=0&amp;Board=exchange&amp;Number=4011389&amp;p age=0&amp;fpart=all):

This is a popular question in freshman physics. Imagine this set-up (because I don't want to take the time to MS Paint it): You have a wall at the front end of a cart and you're standing on the back end of this cart facing the wall and you have a bunch of perfectly elastic tennis balls. You throw the balls against the wall and they bounce off. Describe the motion of the cart, assuming perfect bearings and a smooth, no-slip condition between the flat ground and the wheels.

If you can't figure out what I'm talking about, imagine Daryn's fan on skateboard picture and replace the fan with a little person throwing tennis balls against the "sail".

[/ QUOTE ]


I assume you mean the wall never deforms, there is no friction or drag of the cart apparatus interacting with the ground, etc. such that this is a pure first-approximation problem.

Further assume the ball thrower both throws and catches the ball accurately every time, and that he throws such that the ball hits level with his throw.

In that case, the cart herky-jerks but the net is it moves forward - quite quickly after enough throws, unless you assume there is no air resistance, in which case it herky-jerks but does not move.

Bonus question: Does gravity affect the motion of the described ball-and-cart?

[Patrick del Poker Grande]

I knew this one would donk. Woe is me... back to the plane thread!

[swede123]

Here's what will happen in your scenario.

http://i9.photobucket.com/albums/a71/jonasola/speedingcart.jpg

Swede

[asofel]

why is he cumming out of his hand?

[Freakin]

This seems really simple. You throw a ball towards the wall, you move away from the wall. Am I missing something? The balls being elastic doesn't matter. If you were to catch the balls when they bounced off the wall, you would have additional force moving you away from teh wall

[codewarrior]

What do you mean? An African or European swallow?

[Senor Cardgage]

[ QUOTE ]
This seems really simple. You throw a ball towards the wall, you move away from the wall. Am I missing something? The balls being elastic doesn't matter. If you were to catch the balls when they bounced off the wall, you would have additional force moving you away from teh wall

[/ QUOTE ]

The wall is attached to the cart.

[slickpoppa]

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Edit:In the brief time between when you release the ball and when the ball hits the wall, you travel in the -x direction.

[/ QUOTE ]

scratch what i just said

[mmbt0ne]

[ QUOTE ]
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Edit:In the brief time between when you release the ball and when the ball hits the wall, you travel in the -x direction.

[/ QUOTE ]

Thats not true, unless you are standing on roller skates or something. When you throw the ball in the +x direction you do not have any momentum in the -x direction because that momentum is absorbed by the friction between your feet and whatever you are standing on.

[/ QUOTE ]

You're close, but you need to take one more step to get there. /images/graemlins/smile.gif

[MrWookie47]

[ QUOTE ]
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Edit:In the brief time between when you release the ball and when the ball hits the wall, you travel in the -x direction.

[/ QUOTE ]

Thats not true, unless you are standing on roller skates or something. When you throw the ball in the +x direction you do not have any momentum in the -x direction because that momentum is absorbed by the friction between your feet and whatever you are standing on.

[/ QUOTE ]

I believe we're standing on the cart, and said cart has perfect bearings. So, yes, we do move.

[goofball]

When I first read the question I understood it to be the cart is next to the wall but seperate from it. You also make it unclear whether or not we catch the balls after we throw them. This leaves 4 cases.

Case 1: Detached wall, we don't catch the balls

As you threw the balls off you and the card would move in the opposite direction under the laws of conservation of momentum. You would gain speed m(ball)/m(cart)*vball for each ball.

Case 2: Detached wall, catching the ball.

You would gain the above velocity each time you threw a ball, and again each time you caught one.

Case 3 Attached wall, not catching the ball.

You would move opposite the direction if your initial throws. The ball bounces off the wall and heads off in the other direction, behinid you. You will move opposite the final motioni of the ball.

Case 4 Attached wall, catching the ball

After you do everything you will be again sitting on the cart with a bunch of motinoless balls. You must also be motionless.

[CrazyEyez]

[ QUOTE ]
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Edit:In the brief time between when you release the ball and when the ball hits the wall, you travel in the -x direction.

[/ QUOTE ]

Thats not true, unless you are standing on roller skates or something. When you throw the ball in the +x direction you do not have any momentum in the -x direction because that momentum is absorbed by the friction between your feet and whatever you are standing on.

[/ QUOTE ]

You're close, but you need to take one more step to get there. /images/graemlins/smile.gif

[/ QUOTE ]
Friction keeps your feet from moving on the board but that energy is transferred down to the wheels and, since perfect bearings, the board moves backward. (Then it stops when the ball hits the wall.)
Right?

[mmbt0ne]

[ QUOTE ]
Friction keeps your feet from moving on the board but that energy is transferred down to the wheels and, since perfect bearings, the board moves backward.

[/ QUOTE ]

Exactly. Think about having 2 blocks, one velcro'd on top of the other, sitting on a sheet of ice. If you push the top block, the whole system slides.

[CrazyEyez]

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Friction keeps your feet from moving on the board but that energy is transferred down to the wheels and, since perfect bearings, the board moves backward.

[/ QUOTE ]

Exactly. Think about having 2 blocks, one velcro'd on top of the other, sitting on a sheet of ice. If you push the top block, the whole system slides.

[/ QUOTE ]
Ha. I'm not stupid, despite my efforts to prove otherwise in the plane thread.

[mmbt0ne]

+ is right and up.

[ QUOTE ]
Case 1: Detached wall, we don't catch the balls

As you threw the balls off you and the card would move in the opposite direction under the laws of conservation of momentum. You would gain speed m(ball)/m(cart)*vball for each ball.

[/ QUOTE ]

Correct.

http://www.f2f2s.com/images/case1.jpg


[ QUOTE ]
Case 2: Detached wall, catching the ball.

You would gain the above velocity each time you threw a ball, and again each time you caught one.

[/ QUOTE ]

Correct.

http://www.f2f2s.com/images/case2.jpg


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Case 3 Attached wall, not catching the ball.

You would move opposite the direction if your initial throws. The ball bounces off the wall and heads off in the other direction, behinid you. You will move opposite the final motion of the ball.

[/ QUOTE ]

Incorrect. You'd remain still.

http://www.f2f2s.com/images/case3.jpg


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Case 4 Attached wall, catching the ball

After you do everything you will be again sitting on the cart with a bunch of motinoless balls. You must also be motionless.

[/ QUOTE ]

Incorrect, you'd move in my + direction.

http://www.f2f2s.com/images/case4.jpg

[MonkeeMan]

Why throw tennis balls?

http://www.musicavirtuale.net/mondovip/mariasharapova/images/modelling3_jpg.jpg

[goofball]

In cases 3 and 4 you are complicating things too much, think about them simply in terms of initial and final momentum of the system. They have to be the same right? (let's take the case of just one ball, it's simpler and the same)

In every case the system starts with p = 0 (p is momentum). So in every case it must end with p = 0. In case 3, the final state has the tennis ball going backwards with some velocity x, since the tennis ball is part of the system p(tennis ball)+p(rest of system) must = 0. Since p(tennis ball) isn't zero then p(rest of system) can't be either. It must be equal and opposite.

Case 4 is the same question except at the end of the experiemnt the tennis ball isn't moving, so neither can the cart/person. In case 4 what you are claiming is that our system is going to have a spontaneous increase in momentum without the introduction of an outside force. Doesn't that strike you as odd.

[MelchyBeau]

why aren't these threads in SMP?

Melch

[Patrick del Poker Grande]

[ QUOTE ]
why aren't these threads in SMP?

Melch

[/ QUOTE ]
Meh. I don't like The Trainwreck and Ray put the original (plane) thread in here. Plus, it's much more fun in here. I would seriously hope that the plane thread wouldn't have made it past one page in The Trainwreck.

[bobman0330]

You're wrong about the force the wall puts on the ball. It's -2 x f if f = force imparted by you throwing the ball.