Sandbag Experiment: Reason Why Some Believe Motion 0m/s, Others 10m/s

  • Thread starter Thread starter Beretta
  • Start date Start date
  • Tags Tags
    Experiment
AI Thread Summary
The discussion centers on the motion of a sandbag dropped from a balloon ascending at 10 m/s. Some believe the sandbag falls straight down with an initial speed of 0 m/s, while others argue it retains the balloon's upward speed of 10 m/s momentarily. The kinetic energy perspective suggests that the sandbag, having been in motion with the balloon, initially continues moving upward before gravity acts on it. Observers inside the balloon perceive the sandbag as stationary relative to them, while external observers see it moving upward before it begins to decelerate and fall. Ultimately, both viewpoints are valid depending on the frame of reference considered.
Beretta
Messages
38
Reaction score
0
There is a sandbag dropped from a balloon. At the moment the sandbag is dropped, the balloon is moving upwards at 10 m/s. Why would some people believe that the motion of the sandbag is straight downward with the initial speed 0 m/s, and other think that the motion of the balloon goes up after it is dropped with an initial speed 10 m/s. Give as many reasons as possible for each point.

I know that the balloon would go up with an initial speed of 10m/s that is the balloon speed and then would falls down, but what reasons should I give more then the sandbag is following the balloon's speed?
 
Physics news on Phys.org
I don't exactly understand what you are asking but I'll answer what I think you asked.


The kinetic energy approach:
When the sandbag was connected to the balloon it had a velocity. Since the sandbag has mass, it also has a kinetic energy. Before the sandbag can start falling, the kinetic energy must be brought to 0 (which also means 0 velocity). The upward distance traveled by the bag before it starts to fall can even be calculated by making the work done by gravity equal to the initial kinetic energy of the bag:

Fd = \frac{mv^2}{2}

mgd = \frac{mv^2}{2}

gd = \frac{v^2}{2}

d = \frac{v^2}{2g}
 
Originally posted by ShawnD
I don't exactly understand what you are asking but I'll answer what I think you asked.


The kinetic energy approach:
When the sandbag was connected to the balloon it had a velocity. Since the sandbag has mass, it also has a kinetic energy. Before the sandbag can start falling, the kinetic energy must be brought to 0 (which also means 0 velocity). The upward distance traveled by the bag before it starts to fall can even be calculated by making the work done by gravity equal to the initial kinetic energy of the bag:

Fd = \frac{mv^2}{2}

mgd = \frac{mv^2}{2}

gd = \frac{v^2}{2}

d = \frac{v^2}{2g}


So what u'r referring is on the conservation of energy:

PE=KE
mgh=1/2mv2
 
Sandbag

I need to explain yhy would some people believe that the motion of the sandbag is straight downward with the initial speed 0 m/s, and other think that the motion of the balloon goes up after it is dropped with an initial speed 10 m/s. And I have to give as many reasons as possible for each point.
 


Originally posted by Beretta
I need to explain yhy would some people believe ...
Do you believe that one of these two groups is incorrect? I think that they are both correct, given appropriate context. Here's something that may help: motion is relative.
 
Sandbag

Are you telling me that if you drop a sandbag from a balloon that is going up with a speed of 10m/s then the sandbag would fall down right away with an initial speed of 0m/s? I believe the sandbag would go up a bit with an intial speed of 10m/s which is the initial speed of the balloo and then falls down. Nevetheless, I am ready to listen to your point of view.
 


Originally posted by Beretta
I need to explain yhy would some people believe that the motion of the sandbag is straight downward with the initial speed 0 m/s, and other think that the motion of the balloon goes up after it is dropped with an initial speed 10 m/s. And I have to give as many reasons as possible for each point.
Speeds change according to a=f/m. If its going up when dropped, Newton's first law says it must take time to decelerate.
 
sandbag

I understand all that, but what "reasons" (with an s) should I give for each viewpoint?? I know that by Newton second law the sandbag would take time to decelarate.
 
.
Originally posted by Beretta
I need to explain yhy would some people believe ...

I have no way of explaining why some people believe one thing and others another. This is a Physics forum, not a PSYCHICS forum. We can tell you what happens in a given physical situation but not what someone thinks happens
 
  • #10
sandbag

OK tell what is happening in each situation please!
 
  • #11


Originally posted by Beretta
Are you telling me that if you drop a sandbag from a balloon that ... the sandbag would fall down right away with an initial speed of 0m/s?
In the balloon frame, why not?




Originally posted by Beretta
I believe the sandbag would go up a bit with an intial speed of 10m/s which is the initial speed of the balloo and then falls down.
I would find it quite strange if, when I let go of an object, it shot upwards at 10 m/s. You can choose whatever frame you want. The two most popular ones in this question are probably both accelerated, and they happen to be moving at 10 m/s with respect to each other. But neither one is more valid than the other.

If I see the balloon wiz by me at 10 m/s, then, to me, the sadbag is already wizzing by at 10 m/s. It doesn't have to be dropped before it can have this motion.

If I've got a sandbag in my balloon, then it is stationary. Immediately after I drop it, it is still stationary because it has inertia (tends to stay at rest).

You should not infer (from the situation as stated) that one of these two frames is preferred/more appropriate. In fact, it is just as valid to say that the bag is initially traveling at 0.5c, in the appropriate frame.
 
  • #12
sandbag

The balloon initial speed is 10m/s, and the sandbag, while is still in the balloon, still has the same speed 10m/s. If I drop the sandbag from the balloon I would see the sandbag going straight downward with an initial speed 0m/s. In fact, I think this perception would happen only to those standing inside the balloon. Moreover, for those who are watching the balloon from a distance they would see the sandbag traveling with the same balloon speed at 10m/s for some fraction of time and then right away after dropping the sandbag would start decelerating its “initial speed”. I didn’t mean the when dropping the sandbag it would shot upwards at 10m/s. Since the balloon is still moving up at 10m/s and the sandbag (at the moment of drop) was “immediately” separated from the force that is pulling the balloon up, the balloon would keep traveling upward and sandbag would be travel downward and that what may lead those inside of the balloon to for the first reasoning. On the other hand, from the spectator point of view the sandbag would keep going up at “initial” speed of 10m/s the would immediately starts decelerating after the drop took place and therefore it would travel upward a certain distance before a freefall would take place whose effected by its weight. Please correct me if I am wrong.
 
  • #13


Originally posted by Beretta
Please correct me if I am wrong.
I would agree with everything you said except ...




Originally posted by Beretta
it would travel upward a certain distance before a freefall would take place ...
Freefall takes place the moment it looses the influence of all forces but gravity. Neglecting air resistance, it would experience freefall the moment it left the balloon, regardless of the magnitude or direction of its initial speed.
 
  • #14
The basic rule here is

\frac {d^2x} {dt^2} = -g

with
\frac {dx} {dt}_{t=0} = 10 \frac m s

x(0) = H , where H is the height of release.

This has the solution
x(t) = -\frac 1 2 g t^2 + 10 t + H

The balloon has equation of motion

x(t) = 10t + H

These can be plotted to see exactly what the sandbag does.

Since the occupants of the balloon are experiencing constant velocity, they will observe the bag to fall just as you do when you drop an object while standing in your room.
 
  • #15
sandbag

The balloon initial speed is 10m/s; meanwhile, the sandbag while is still in the balloon, has the same speed of 10m/s; at that point, we can think of the balloon and the sandbag as one object. If for instance the sandbag has been dropped from the balloon, those standing in the balloon would feel the sandbag going straight downward with an initial speed 0m/s because in fact the relative speed of the sandbag to them is 0m/s. Furthermore, at the first moment of drop taking place (the first moment of the sandbag is in the air), the sandbag is on its own with no fed upward speed from the balloon. At that very moment those in the balloon are continuing of traveling upward with the speed of 10m/s watching the sandbag getting more distant from them. In fact, this perception would happen only to those standing inside the balloon.

On the other hand, for those who are watching the balloon from distance (from a spectator point of view) would see the sandbag traveling with the same balloon speed at 10m/s for some fraction of time (initial speed) after it has been dropped. Then immediately after the dropping took place, the sandbag would start decelerating its “initial speed” of 10m/s. Decelerating its initial speed? Yes, then it’s still traveling upward since it has a speed. An object in motion continues to travel with constant velocity unless acted on by an external force – Newton first law. So, what external force is acting on the sandbag to change its continues traveling with its initial speed (regardless to outside factors like air resistance)? Mass of course. At that moment a battle is taking place between the mass of the sandbag (decelerating it and pulling it downward) and its velocity (pulling it upward). The moment the mass wins – speed is 0m/s; a fall would take place. For sure the middle picture is the correct one from an experimental point of view.


I was thinking of another example that may describe, with a certain similarity, the same situation. Once I was sitting next to my friend who was driving a car. At that point I wasn’t looking at the speedometer, but let's just assume it was 50km/h. We were lazy to stop the car next to a trash container, so I could drop my soda can. So why not shifting the car’s direction to get closer to the container and throw the can without stopping the car my friend says, and that exactly what happened. The car starts taking the right lane direction long ago before reaching the container, and then I threw the can out of the window when the container appeared. What happened is the can, unlike what I assumed would happen, not only missed the container but also kept traveling a certain distance on the ground. Another example is warplanes bombs. Warplanes have to release their bombs before the plane reaches its target. Like in the World War II when the cam is recording the falling bombs from inside the plane ground hole, you can still see the bombs traveling with the plane for a certain time after their release. One more example is the space rocket. At a certain point of traveling the space rockets has to release its lower part. Those sitting in the space rocket for instant would see the lower part of the rocket falling instantly at an initial speed 0m/s. Meanwhile, when the event is broadcasted on TV, we can clearly notice that the lower part would keep rising upward following the space rocket before finally starting to fall, and that exactly what happened in the balloon and the sandbag.

----------------------------------------------------------------------

Is it more complete now or is there still something I missed?
I would like to thank you all for your guidance.
 
Last edited:
Back
Top