-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
  • 220,844
PhysicsNoob
  • 1

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
  • 27,443
-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
  • 2,374