# Conservation of Momentum Question

1. ### crddrc

2
Hello,

Here is the situation I have been pondering:

A person is sitting in a shopping cart at rest. By throwing their weight forward they are able to cause the shopping cart with them inside to move. My question is how this works. If momentum is conserved unless an external force acts on the system and the initial momentum is zero when the cart is at rest, how is the shopping cart with the person inside able to move forward?

Thanks

2. ### K^2

2,470
If the cart is completely frictionless, it won't work. The center of mass will stay put.

In the real world, however, there is going to be friction. You can use static friction to allow you to pick up the speed without pushing the card back. Effectively, via friction of cart with ground, you are pushing from the ground.

Once you pick up speed, you hit the wall of the cart. The impact generates brief, but very high force. That force easily overcomes static friction causing the cart to roll. Once in motion, friction is much lower, so you can travel some distance before coming to a stop.

In terms of conservation of momentum, the momentum of you + cart is not conserved, but that's not a closed system due to friction. Momentum of you + cart + Earth is conserved.

3. ### Simon Bridge

15,261
Welcome to PF;
Notice that the cart moves in the direction the person throws themselves? Throwing forward should make the cart go backwards shouldn't it? That should be a clue... there is friction involved.

It is the sudden shift from going forward to stopping that overcomes the static friction, allowing the cart to move. The change in momentum in that short time is backwards, so the cart goes forwards. (You may not think of this as an "impact" ;) )

4. ### crddrc

2
Thank you Simon and K^2! Great answers from both of you.

5. ### cragar

Is this similar to sitting in the cart with a paint ball gun and shooting the front of the cart.
Because I thought if we were shooting the front of the cart it would go forward because the change in momentum is greater when the ball bounces off, even with no friction.
Or I guess kicking might be different, Im probably way off.

6. ### jbriggs444

1,847
[Assuming the wheels are frictionless...]

If the paint ball lands inside the cart then momentum is conserved and you end up motionless.

If the paint ball lands outside the cart then what you have is essentially a needlessly complex rocket motor.

7. ### Matterwave

3,859
I should mention that in a frictionless case, the cart would move backwards while you were running forward and then perhaps go forward a bit when you hit the front, leaving the COM at the same place. But because running forward provides a much smaller force, the cart does not overcome static friction until you hit the front of the car.

The possibility of moving the car forward while you're in the car is the same as the possibility of moving yourself forward while standing without anybody pushing you.

8. ### Buckleymanor

554
Figueres if that's the case, would a bow and arrow and a shortish peice of string attached to the bow and the arrow be similar to the OP question.
You sit in the cart and fire the arrow.
Would the cart move backwards when you fire.
Would the cart move forewards when the arrow reached the end of it's flight and jolted the string, the bow, the person, the cart.

9. ### K^2

2,470
Frictionless case? The cart would recoil and start moving backwards when you fire the arrow. The cart and arrow would come to a stop once the string is fully extended. The center of mass would remain stationary throughout the experiment.

10. ### Simon Bridge

15,261
... when you reel in the arrow, slowly, the cart stays in place. Reel in the arrow quickly and you risk overcoming the static friction.

11. ### A.T.

6,173
Aside from the static friction that prevents rolling, there is also the lateral static friction that prevents wheels from sliding sideways. Since a shopping cart has caster wheels this also plays a role when someone is throwing his weight around in the cart. Even with wheels that have zero static rolling friction, you could potentially get the center of mass moving.

12. ### Buckleymanor

554
Why would you start moving backwards you the bow and the cart would recoil but I don't imagine it's in that direction.

13. ### K^2

2,470
The movement of limbs actually results in two recoils. When the limbs start to accelerate forward to propel the arrow, the bow actually pushes towards you, but it's a fairly gentle push over longer time. When limbs stop, it's rather sudden because of the string, so you get the kick that the quote talks about. It's particularly severe on recurve bows simply due to the geometry of the limbs. You feel that kick a lot more because of how sudden it is, but the actual momentum transfer is smaller, because the initial push contains both momentum transferred to the limbs and to the arrow, and the arrow's momentum is not recovered.

It's the net difference due to the arrow's momentum that results in your total momentum after firing the arrow being directed backwards. So the cart will roll backwards as result of the recoil until the arrow is stopped by the rope tied to it.

Edit: This would actually be a fun one to demonstrate like that. If the weather improves for a day or two, I'll try to film it. The momentum carried by the arrow is quite significant. Should be enough to get me rolling on a skate board.

14. ### Buckleymanor

554
Not so sure that the bow pushes much towards you once you let the arrow go. It's in tension as you draw the bow and is pushing towards you as you hold it.Once you let go it can only move towards you with the same force as the arrow and string is moveing in the opposite direction.In all there must be three recoils two in one direction and one in the other.
My bet is the two recoils has more force than the one.
Go for it and let us know the results.

15. ### jbriggs444

1,847
Since momentum is strictly conserved and the arrow ends up at rest, my bet is that the impulse delivered by all of the recoils plus the tug on the string attached to the arrow sums to zero.

16. ### K^2

2,470
It is basically your guess against conservation of momentum. Doesn't look good for your bet. :)

But I've got nothing against the experiment. So I'll set it up. I've figured out location, so now I just need to borrow a skate board from somebody. (Mine's a 2-wheel caster, so that won't work.)

Now, what would be really great is if you could see the influence of all the individual recoil stages, but I doubt I'd be able to catch that on camera. The net result should be plainly visible, though.

17. ### cragar

ok when I fire the paint ball into the front of the cart it will move forward right
becuse the change in momentum is greater when it bounces of the front of the cart .
ok so now it is headed to the back of the cart and it bounce of their so that should appose the motion. Now its headed back to the front of the cart. it still seems like the cart will move forward on average. There is porbably some subtle things I am missing.

18. ### rcgldr

7,561
If you only consider the cart, then the external force is geneated by the earth in reaction to the internal force which is transmitted to the earth via friction between the wheels and the earth (Newton third law pair of forces). Even if the wheels are aligned in the direction of movement, friction in the axles of the wheels is enough for a small reaction force related to the person moving within the cart to be applied over a longer period of time, then overcome by a sudden jerk by the person in the opposite direction.

If you consider the cart and earth as the closed system, with the car initially at rest with respect to the earth, then the center of mass of cart, person, and earth does not move, regardless of any relative movement between cart and earth, and momentum of this system is conserved.

19. ### Simon Bridge

15,261
Conservation of momentum says that the recoil has to push the cart backwards - the bet here is whether the initial backwards push is better than friction in the wheels?

20. ### K^2

2,470
This is a bit difficult to set up with cart that's also moving, but if we assume that cart is "infinitely heavy", you can do an estimate.

Say, after each bounce, velocity goes from v to -av for some a 0<a<1. Technically, we'll get infinitely many bounces in this approximation. But this kind of infinite sum we can still do.

So first, the ball is fired, and cart receives -mv. When the ball hits front wall, ball recoils with -av, and cart receives (a+1)mv. Now the ball bonces off from the back wall, boing +a²v now, and cart picks up -(a²+a)mv.

Total transfer: Δp = -mv + (a+1)mv - (a²+a)mv + (a³+a²)mv - ...

Rearranging terms slightly: Δp = -mv + mv + amv - amv + a²mv - a²mv + a³mv - ...

So every term cancels at except for the last one. After infinitely many bounces $\Delta p = \lim_{n \to \infty} (-1)^{(n+1)}a^nmv = 0$.

The shortcut, of course, is to note that if the final velocity of the ball relative to cart is zreo, then so was the total momentum transfer. Of course, like other people have said, friction can make a difference.

P.S. I've tracked down a skateboard I can borrow and will set everything up in the next couple of days. I'll try to see if I can manage to arrest the arrow with a string as well, but it might be a bit difficult to achieve.