-1

I just learnt the concept of conservation of momentum, and wanted to make sure my thinking was correct on a toy example.

Question

Suppose you're in a car full of frictionless sand traveling on a frictionless road at constant speed $v$. A hole suddenly appears in in the bottom of the car, and the sand starts pouring out. My question is does the car speed up as the sand pours out?

Reasoning

My thinking is that for the the system of (myself, the car, and all the sand) the momentum is not conserved, because the sand that falls out gains vertical momentum from falling.

If we choose the system as (myself, the car, and the sand still in the car), the momentum is conserved, because the external forces in the vertical direction (the normal forces and the weights) cancel out, and there are no external forces in the horizontal direction.

Therefore, we have the initial momentum equalling the final momentum: let $m$ be my mass, let $m_c$ be the mass of the car, and let $m_s$ be the mass of the sand in the car still remaining.

We have $$(m + m_c + m_s)v = (m + m_c + m_s)v_f,$$ so clearly $v_f = v$.

Is this logic correct? Would love some tips/clarification on picking the system and determining the momentum if not.

Qmechanic
  • 201,751
PhysicsNoob
  • 1
  • 1
    The leaking cart is a bit more complicated than that. See https://physics.stackexchange.com/q/1683/25301, https://physics.stackexchange.com/q/659460/25301 along with the inverse problems https://physics.stackexchange.com/q/699160/25301, https://physics.stackexchange.com/q/153361/25301 and so on – Kyle Kanos Aug 24 '23 at 17:09
  • @KyleKanos Very interesting. If I simplify the problem so that the sand doesn't leak out, but the cylinder of sand above the hole all falls out at once, leaving the rest of the sand around the hole untouched, we do have that the velocity stays the same, yes? And thanks for the response! – PhysicsNoob Aug 24 '23 at 17:23
  • An important feature of momentum, unlike energy, is that it is a vector quantity - if we look at the vertical and horizontal directions (as you did), there is horizontal momentum and vertical momentum, and they are separate. The sand gaining vertical momentum, as you noted, will not change the horizontal momentum (and thus the horizontal velocity) at all. But as Kyle noted above, this problem is more complicated - the sand not only moves vertically out of the hole - the sand in the container will also shift horizontally to "even itself out", and can change the horizontal momentum. – Nadav Har'El Aug 24 '23 at 17:39
  • @Nadav Har'El Great explanation, thank you! – PhysicsNoob Aug 24 '23 at 17:55
  • @PhysicsNoob In either case (slow leak or fast leak), you have a variable mass system in which the conservation of momentum must be handled very carefully. – Kyle Kanos Aug 24 '23 at 18:02
  • Well I know the car does not accelerate, not even if the hole is at the rear of a long car, or at the front. Zero momentum transfer means zero force means zero acceleration. – stuffu Aug 24 '23 at 18:13
  • Momentum is still conserved, except now its $\Delta (mv)$ instead of $m\Delta v$. – Bob D Aug 24 '23 at 18:18
  • One also has to scrutinise the physical interaction. The actual location at which the horizontal momentum of the sand changes is when it interacts with the floor. There is nothing communicating that change back onto the car. So the car should not be changing velocity. This is completely different to the case whereby sand is dropped from above into the car. – naturallyInconsistent Aug 25 '23 at 02:28

2 Answers2

0

The sand falling has zero vertical momentum when it leaves the car, so no vertical momentum change occurs. The sand does have horizontal momentum, so the horizontal momentum of the car plus the remaining sand changes. However this change is purely due to the change of the mass and the velocity remains constant.

I cheated for the vertical momentum. The car is vertically accelerated by gravity and the reaction force of Earth. As the falling sand does not feel the reaction force, there is a reduction of reaction force of the car plus the remaining sand, exactly compensating the reduction in gravity. Now you may take the tyre pressure and the suspension system in consideration, and the spring constant of the tarmac. These combined springs relax ignorer to reduce the reaction force so the car will move up a little. Unless is is a hydraulically suspended one.

Let's ignore the centrifugal force due to Earth ...

my2cts
  • 24,097
-1

Your equation makes no sense. Is $m_s$ the same number at both sides? Is the left side of the equation a situation at different time than the right side? If so, then the $m_s$ should be a different number at the left side and the right side.

Anyway, there is nothing to calculate.

Think like this: There is a car. Is there momentum transfer between the car and something else? No. So the car does not accelerate.

By 'car' I mean a thing consisting of wheels, motor and chassis.

stuffu
  • 1,978
  • 11
  • 11
  • Is there momentum transfer between the car and something else? No. So the car does not accelerate. I suggest you read through the links I posted as a comment to the original post as this block I've quoted is definitely wrong. – Kyle Kanos Aug 24 '23 at 21:00
  • @KyleKanos "The water jet from the nozzle is directed vertically down." The sand flow is NOT directed vertically down, as there is no adjustable nozzle in the car. Seems like two very different things to me. Did not read everything. – stuffu Aug 25 '23 at 03:07
  • From OP: A hole suddenly appears in in the bottom of the car, and the sand starts pouring out If the bottom is not pointed vertically down, then we have very different understandings of either vertical or bottom. Either way, as I commented to the OP: you have to be very careful in how you treat conservation of momentum. This is because the mass that is being lost (the sand) actually contributes to a slight gain in the speed of the car ($\propto v\dot{m}$)! – Kyle Kanos Aug 25 '23 at 10:59