Suppose a wagon is moving at constant velocity on a friction-less surface, and rain begins to fill the wagon.

The net force on the wagon is zero, so momentum is conserved; as the mass of the wagon increases, the speed decreases. But if the velocity of the wagon changes, the net force can't be zero, right?

There has to be some force opposing the motion of the wagon to slow it down. How do we reconcile this?

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    $\begingroup$ What happens to the raindrops falling into the wagon? $\endgroup$
    – CuriousOne
    Commented Dec 15, 2014 at 3:04

6 Answers 6


To deal with this type of problem, you must be careful to define exactly what system you are dealing with, and then not change that system part way through the problem. This definition allows you to be very clear about whether the "system" has any external forces acting, and thus whether the momentum of the system is constant or not.

In this case, you seem to be defining the wagon itself as the system, but then talk about the wagon as gaining weight, implying that the definition of what constitutes the wagon system is changing.

Let's try this: the system is the wagon itself, without any stray mass that may be added. The wagon has a certain amount of momentum, and since there is no outside force of friction, that momentum is constant.

But then the rain starts to fall. None of this rain is included in the system, even though it gets trapped inside the wagon. But in being trapped, the vertically falling rain also exerts an horizontal force on the system: either impacting the back of the wagon in the air, or hitting the bottom, and flowing towards the back of the wagon. All this means that there is an external force exerted by the rain on the system, and momentum of the system is not conserved.

We can start over: the system now is defined as including the wagon and all the vertically falling water. Since the rain initially has no horizontal velocity, the total momentum of this new system is just that of the wagon.

Now the rain starts hitting the wagon. But under our new definition, all of the rain impacting is an internal force, and cannot change the total momentum. This new system is isolated and momentum is conserved. So now, since more and more of the system is travelling with the wagon, the wagon must slow down. Internally, momentum is being transferred from the wagon part to the rain part of the overall system.

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    $\begingroup$ +1 for focusing on the definition of the system - the most common source of confusion in classical physics! $\endgroup$
    – thomij
    Commented Dec 15, 2014 at 16:37
  • $\begingroup$ Brilliant explanation. However, in the vertical direction, taking the system to be the rain and wagon, momentum is not conserved: y momentum before = mass of rain x y component of rain and momentum after = mass of rain x zero because the y velocity of the rain after falling into the wagon is zero. The wagon has no y component of velocity so it doesn't feature in this equation. So how is it that momentum is not conserved when you take the system to be the rain and wagon in the y direction? $\endgroup$
    – Amit
    Commented May 10 at 15:26

When the raindrops hit the wagon's surface, they aren't moving relative to the tracks. Friction is required to accelerate the raindrops to the wagon's speed. By Newton's third law, there must therefore be a reaction force on the wagon surface by the raindrops.

  • 2
    $\begingroup$ What if the raindrops happen to be falling with a horizontal velocity exactly equal to that of the wagon? $\endgroup$
    – Benjam
    Commented Dec 15, 2014 at 7:47
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    $\begingroup$ @Benjam If the raindrops are falling at the same velocity as the wagon, then the wagon does not slow down. The mass of the wagon is increasing, but also the momentum is increasing: each raindrop is adding its momentum to the wagon. $\endgroup$
    – mpv
    Commented Dec 15, 2014 at 9:58
  • 2
    $\begingroup$ Would you really say that friction is required to accelerate the raindrops? If a raindrop falling into the wagon slid frictionlessly to the back, would it not be pushed by -- and therefore push back against -- the rear wall of the wagon? $\endgroup$
    – AmigoNico
    Commented Dec 16, 2014 at 5:57

The mass of the cart is changing! This is the variable-mass system, which says, $$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ where $v_{rel}$ is the relative velocity of the escaping/entering mass. In your case, there are no external forces so, $$ v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$ So the change of velocity comes from the change in the mass.


But if the velocity of the wagon changes, the net force can't be zero, right?

Only true if the mass is constant (it's not if a wagon is filling up with water). If mass and velocity both change, you can't say anything about the force.

Record the experiment and play the video backwards. You will see a wagon moving backwards. The wagon is spraying water out. Relative to the wagon, the sprayed water appears to be moving upwards and towards the front of the wagon. The wagon is accelerating in the opposite direction of the horizontal velocity of the sprayed water, with the vertical velocity neutralized by the ground. The image is analogous to a rocket that is skidding by pushing into flat ground at an angle.

Consider the classical experiment of a railway cart with a machine gun firing backwards. Angle the gun above the horizon, replace bullets with water, and then reverse time - you have arrived at your wagon experiment.


This is the opposite of the rocket problem. In a rocket, acceleration occurs becaue mass is thrown out the back end.
F = d/dt(mv) = m.dv/dt +v.dm/dt.
If F= 0, then m.(-dv/dt) = v.dm/dt <-- note the negative term with the acceration.
In a "typical" problem mass does not tend to change significantly, but in the rocket this mass term is highly significant. Same with the wagon here, the mass is not decreasing, but increasing.
So, just as our equatation shows, an experimenter would expect to see a "negative" acceleration (-dv/dt) decreasing the velocity of the wagon as the mass increases.


The net force on raindrop plus wagon is zero. Consider a single rain drop.

Let the momentum in the direction of travel of the combined wagon/raindrop system be p. Now

p = p_wagon_before + p_raindrop_before 

where p_wagon_before & p_raindrop_before are the momentum of the wagon and the raindrop before the drop hits the wagon. We have then:

p_wagon_before = p and 
p_raindrop_before = 0. 

After the drop lands in the wagon we'll have

p_wagon_after + p_raindrop_after = p_wagon_before + p_raindrop_before.

Let us assume that p = M V where M is the mass of the wagon and V is it's velocity prior to the collision. After the wagon/raindrop collision we have

p = (M + m) (V+dV) 

where m is the mass of the raindrop and dV is the change in velocity due to the collision so that

M V = (M+m) (V+dV) so that 0 = MdV + mV + m dV so that 

dV = - mV/(M+m) 

So the net force is zero, and the wagon slows down since dV is of opposite sign to V.


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