2
$\begingroup$

Consider a container with lid of total mass, $M$. A mass, $m$, is allowed to fall into the container when the container is open. The whole set up is kept on a sensitive weighing machine and everything is in vacuum. enter image description here

When the mass, $m$, is inside the container the container is closed with the lid not letting any other matter to enter or escape the container.

Neglecting any effect of the movement of the lid, will the weighing machine show the same or different readings during the two stages, when the container is open and when it is closed before the mass $m$ hits the bottom of the container? Why? If there was air instead of vacuum, in the second stage the weighing machine would also measure the weight of the air trapped. But apart from this, would the situation be any different?

$\endgroup$
  • $\begingroup$ The answer is that it depends. What did you do to solve the problem? $\endgroup$ – CuriousOne Jan 2 '15 at 6:05
1
$\begingroup$

While the mass is falling, it is not interacting in any way with the weighing machine. The weighing machine only responds to forces applied to it, and the falling mass $m$ is not exerting a force on it. The presence of the large container mass $M$ does not matter until the small mass makes contact with it so that a force can be transmitted to the weighing machine.

If there is air or other fluid, then the amount of force transmitted depends on the viscosity and how close the small mass is to terminal velocity. For a mass falling through air, there will hardly be any change in the mass reading. For a mass descending at a constant speed in water, the weight reading will be the sum of the large mass, the small mass, and the water column directly above the weighing machine.

$\endgroup$
  • $\begingroup$ What if there is air. Air is a fluid. If we measure the weight of water and some mass immersed in it, it will measure the weight of both water and that mass. $\endgroup$ – Jolie Jan 3 '15 at 6:19
  • $\begingroup$ @scisyhp Edited my answer. $\endgroup$ – Mark H Jan 3 '15 at 14:31
0
$\begingroup$

it should show the same weight, but while keeping the lid, which is assumed to have some mass, the normal force by the weighing machine on the system decreases until the lid stops.

$\endgroup$
  • $\begingroup$ Are you sure? Would you like to change your answer now? $\endgroup$ – CuriousOne Jan 2 '15 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.