Physics problem question

[happyboy]

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Why does mass not matter in determining how long it will take for a ball to fall to the ground. IE: If I drop a .5 kg ball vertically and throw a 2 kg rock horizontally, why do they end on the ground at the same time? Equations not necessary.

 

[SeminoleVesicle]

Potential Energy is converted to Kinetic Energy.

mgh=1/2mv^2

The mass cancels out on both sides.

 

[Temperature101]

Why does mass not matter in determining how long it will take for a ball to fall to the ground. IE: If I drop a .5 kg ball vertically and throw a 2 kg rock horizontally, why do they end on the ground at the same time? Equations not necessary.

The above poster give yu a perfect answer...

 

[MT Headed]

Potential Energy is converted to Kinetic Energy.

mgh=1/2mv^2

The mass cancels out on both sides.

Except that proves something about velocity, and the original question was about time.

The answer, without equations, is that the vertical force of gravity is directly proportional to an object's mass. Heavier objects have a greater force pulling them down than smaller objects. All objects accelerate down at the same rate, no matter how heavy they are, no matter what is going on in the horizontal direction. Objects that vertically accelerate at the same rate, starting with the same vertical speed, will cover the same vertical distance in the same amount of time.

With equations:

F=ma; W=mg; x=1/2at^2

t = sqrt(2x/a) = sqrt(2x/(F/m)) = sqrt(2x/(mg/m)) = sqrt(2x/g)

In that last step, the mass cancels out.

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A Forum.

 

[MD Odyssey]

The previous posters just told you to set potential energy at the top of the cliff to the kinetic energy at the bottom. Since you're asking a rather simple, fundamental question, I'm going to go out on a limb and guess that you probably don't have a solid grasp of energy yet. That's fine - if you don't, you'll get there later.

The way I would say to think about it is the following:

Newton's 2nd law tells us that the sum of the forces acting upon a body is proportional to the acceleration of that body and the constant of proportionality is the objects mass. Which is a somewhat complicated way of saying that:

F = ma

where m is the mass, F is the net force, and a is the acceleration that you want to find. You also know that a is in some way related to the velocity, so once you've found it, you can find the velocity, the time, and so forth using the kinematics equations you've probably seen.

If you start by drawing a free-body diagram, you'll realize that there is only one force acting on the body, namely gravity. The gravitation force most definitely is proportional to the mass. Now, since there is only one force acting on the body, we have the net force (if there were more, we would compute the vector sum first) and then using Newtons equation, we have

-mg = ma

Clearly, in this case, the masses cancel out and you find that the acceleration of the object is just g in the downwards direction. As with the energy arguments given earlier, the mass is irrelevant. The reason this is true in both cases is due to the fact that both the force and Newton's law were proportional to the mass. If for some reason gravity depended upon the square of the mass, this wouldn't be the case.

Hope this helps.

 

[MShopes]

I would see it from other aspect than potential and kinetic energy. Vertically, the equation for distance is X=V0t+1/2 at^2...since the object is dropped, V0=0 and the time will depend on the vertical distance (X) and acceleration only. No mass is involved here in the equation at all.

Horizontally, the equation is X=V0t, or t=X/V0 and in this case, time will depend on the distance and initial velocity. Initial velocity here may or may not be 0 since it is thrown. Overall, the time will depend on these two factors but the mass is not involved at all!

 

[dmplz707]

This brings out an interesting discussion. Now I know this will be completely off topic from the OPs original question, but say for example we have an electric field as opposed to a gravitational field. In the case of the electric field, the masses cannot simply cancel out and will thus have an affect on the objects projectile, is that correct?

 

[MD Odyssey]

This brings out an interesting discussion. Now I know this will be completely off topic from the OPs original question, but say for example we have an electric field as opposed to a gravitational field. In the case of the electric field, the masses cannot simply cancel out and will thus have an affect on the objects projectile, is that correct?

That's a good question. If we were to do this sort of experiment in space where there is no gravity but we have somehow made an electric field, Newton's laws would still hold:

F = ma

Let's give the particle a charge of q. The force on a charge q is just the electric field times the charge, so the net force is

F = qE

We know the net force and can solve for the acceleration directly

a = (q / m)E

Notice that, if we change the sign of the charge, we get an equal and opposite acceleration. Also, I've implicitly assumed an electrostatic and uniform electric field - what I've actually analyzed is the acceleration of a charged particle between two parallel plates, similar to a capacitor. If the field is constant, then the acceleration is also a constant and all the normal kinematics equations you've learned are valid. Easy as pie.

Another thing to notice from that equation is that, if we are neglecting gravity, the acceleration is inversely proportional to the mass. This means that, for two equally charge particles with larger masses, the acceleration for the lighter one would be larger than for the heavier one. This should make sense to you since it's essentially a restatement of the law of inertia.

 

[MD Odyssey]

I would see it from other aspect than potential and kinetic energy. Vertically, the equation for distance is X=V0t+1/2 at^2...since the object is dropped, V0=0 and the time will depend on the vertical distance (X) and acceleration only. No mass is involved here in the equation at all.

What you've said is true, but you've rather missed the point of the OP. Naturally, the kinematic equation for displacement that you used won't include mass - it's derived by assuming that acceleration is a constant and integrating twice. So, in a sense, you've assumed what you wish to prove.

There are plenty of forces for which the mass of the object is important. It just so happens that Newton's law of gravitation depends linearly upon the mass of the falling body and therefore cancels out of the equation. If it depended upon the square of the mass, it wouldn't.

Another thing to point out, just for completeness, is that, implicit in the statement of the 2nd law that we have been using is the assumption that mass is constant. Newton's law is more properly written as the time derivative of linear momentum. I'm not going to go through it here, but since P = mv, taking the time derivative of that quantity will yield two terms if the time derivative of m is non-zero, one of which is proportional to the velocity and another proportional to the acceleration. Solving said equation is well beyond the scope of anything expected to be on the MCAT.