# Question 88a84

Jan 3, 2017

See below.

#### Explanation:

An example of an instance where an object might lose mass as it travels would be a rocket burning fuel. This is obviously on a much larger scale and a rocket has a thrust force to keeping it accelerating in the proper direction, but there are real world problems which have to take this factor of losing mass into account.

Density is given by ratio of mass to volume of an object ($D = \frac{m}{v}$). Just altering the mass will not change the density of the sphere, assuming its still made of the same material. I assume you're only concerned with the mass loss and not particularly a change in density.

I'm going to assume air resistance is negligible for now and examine which properties of the sphere's motion change.

If you halve the mass, of course the force of gravity acting on the object decreases.

${\vec{F}}_{1} = m g$

${\vec{F}}_{2} = \left(\frac{1}{2} m\right) g \implies \frac{1}{2} {\vec{F}}_{1}$

Where $g$ is a constant, the free-fall or gravitational acceleration.

If you double the mass, the force of gravity acting on the object doubles.

Note that this does not affect the vertical acceleration of the object if air resistance is negligible.

If the object suddenly loses half of its mass, its momentum will be suddenly halved at that point also:

${\vec{p}}_{1} = m \vec{v}$

${\vec{p}}_{2} = \left(\frac{1}{2} m\right) \vec{v} = \frac{1}{2} {\vec{p}}_{1}$

And it doubles if you double the mass.

Additionally, suddenly halving the mass also halves the kinetic energy of the object:

${K}_{1} = \frac{1}{2} m {v}^{2}$

${K}_{2} = \frac{1}{2} \left(\frac{1}{2} m\right) {v}^{2} = \frac{1}{2} {K}_{1}$

Again, $K$ doubles if you double the mass.

I believe a sphere specifically would have not only translational kinetic energy but rotational as well, because it would spin as it moves through the air.

${K}_{r o t} = I {\omega}^{2}$

Where $\omega$ is the angular velocity, equivalent to the tangential velocity divided by the radius,

$\omega = {v}_{t} / r$

and $I$ is the moment of inertia, which depends on the object. For a sphere it will depend on whether it is hollow or solid

(2/3mr^2  vs. 2/5mr^2)#.

In spite of all of this, if air resistance is ignored and this is treated as a simple projectile motion problem where there is no horizontal acceleration and the vertical acceleration is a constant $9.8 m \text{/} {s}^{2}$ downwards, I expect nothing about its trajectory will change, because mass has no effect on the motion (velocity, acceleration, flight time). When acceleration is constant (air resistance ignored), we can use basic kinematics to make predictions about the properties of the object's motion ($\vec{v}$, $\vec{a}$, $\Delta t$). We can see that none of these equations include mass:

$\Delta s = {\vec{v}}_{i} \Delta t + \frac{1}{2} \vec{a} \Delta {t}^{2}$

${\vec{v}}_{f} = {\vec{v}}_{i} + \vec{a} \Delta t$

${\left({\vec{v}}_{f}\right)}^{2} = {\left({\vec{v}}_{i}\right)}^{2} + 2 \vec{a} \Delta s$

If air resistance is not negligible, this does make a difference where mass is concerned, but I do not know enough about these effects at this point to teach someone else about them when the effect of the drag force acting on the object is not constant.

Great question! :)