Thread: effects of mass increase of a moving object

in regular earth atmospheric conditions, about sea level,
lets say someone throws an object, that weights 500g,
as hard as he can.

in mid-air, the object starts gaining mass, over.. say 5 seconds, until its got a mass of about 300kg.

will the now 300kg object fly as far, as a 500g object would,
or would it fall short?

i would think for the 300kg object to travel the same distance as the 500g object, it would have to gain energy, as well as mass, but on the other hand,
earths gravity has the same pull on all objects regardless of mass.
so would the object slow down in midair as it gained mass, and slump down from the sky, or would it travel the full distance of a much lighter object?

I would think since the 500g object's velocity is dependant on the energy applied to its 500g mass. when mass is added to the object in flight, its trajectory now would still be determined by the energy applied to the object when it weighed 500g, so I think it would not travel as far as the original object would have if the extra mass had not been added to it.

We are not talking about relativistic effects here are we? If not then the gain in mass is due to an inelastic collision with some other mass, and you have to know the initial velocity of the other mass it collided with to solve the problem.

Conservatoin of momentum, maybe?

The other way to see this is to realize that generally the projectile motion equations are deived using a constant mass. Because it is constant, it appears on both sides of, say, the Euler-Lagrange equations and divides out. If the mass is changnig, it can't divide out on both sides.

Conservation of ENERGY. Stupid. Not momentum. This has been bugging me since I made the post.

No, Ben, it's an inelastic collision so energy is not conserved. Momentum always is though.

Well, energy is conserved of course but it is lost to friction heating so it doesn't figure into the ballistics.

Is the reverse subject to the same forces? An object in flight is more likely in the real world to lose mass....does it then travel farther or the same distance as when it was launched with the greater mass? If an artillery shell was launched covered in ice and the ice melted off half way through the flight, would the shell go further without the ice? (assuming the aerodynamics are not altered).

Originally Posted by Jellyologist

Is the reverse subject to the same forces? An object in flight is more likely in the real world to lose mass....does it then travel farther or the same distance as when it was launched with the greater mass? If an artillery shell was launched covered in ice and the ice melted off half way through the flight, would the shell go further without the ice? (assuming the aerodynamics are not altered).

Okay, let's say in the first case you had a cold object and it's accumulating ice as it moves through the air. The moisture has no initial velocity as it is in still air, so when it attaches to the object, it doesn't add any momentum. The initial momentum is m1v1, the mass of the object multiplied by its velocity. The momentum of the object plus the ice (m1+m2)v2 is also equal to m1v1 by conservation of momentum so v2=v1/(m1+m2) and it slows down.

In the second case, the ice is melting off the object as it flies through the air. So, it leaves the object with the same speed. There is no exchange of momentum. Both objects have the same speed. Before the melting you had (m1+m2)v1 and after you have m1v1+m2v1. This is neglecting any effects of air resistance.

Originally Posted by Harold14370

Originally Posted by Jellyologist

Is the reverse subject to the same forces? An object in flight is more likely in the real world to lose mass....does it then travel farther or the same distance as when it was launched with the greater mass? If an artillery shell was launched covered in ice and the ice melted off half way through the flight, would the shell go further without the ice? (assuming the aerodynamics are not altered).

Okay, let's say in the first case you had a cold object and it's accumulating ice as it moves through the air. The moisture has no initial velocity as it is in still air, so when it attaches to the object, it doesn't add any momentum. The initial momentum is m1v1, the mass of the object multiplied by its velocity. The momentum of the object plus the ice (m1+m2)v2 is also equal to m1v1 by conservation of momentum so v2=v1/(m1+m2) and it slows down.

In the second case, the ice is melting off the object as it flies through the air. So, it leaves the object with the same speed. There is no exchange of momentum. Both objects have the same speed. Before the melting you had (m1+m2)v1 and after you have m1v1+m2v1. This is neglecting any effects of air resistance.

Thanks. I appreciate the formula. You've inspired me to get my pencil out and work through a problem.

I see I left out an m1 in that equation.

v2=v1/(m1+m2)