Buckleymanor said:
I don't see how that will prevent the lighter object reaching the other side of the ship more quickly.
The ship would have to accelerate at the same velocity for both which it won't becuase it's more massive with one object on board than the other.
Ok, let's use some actual formula's for this. Suppose a rocket with an object inside had been constantly accelerated and then instantly reversed the direction of acceleration. At the moment that the acceleration changed, it had a velocity v_0, and we label the location as the origin. The distance the base of the rocket travels at a time t later with new acceleration -a_ris:
d_r = v_0 t - \frac{1}{2} a_r t^2.
Now, up till now, the object was in contact with the base of the rocket, so it also had a velocity of v_0. However, once it loses contact with the ship, it will keep traveling at that velocity until it makes contact again. So the distance that the object travels in the same time t is simply:
d_b = v_0 t.
If the chamber where the object is had a length l, and t was the time it took for the object to travel to the other end, then the object would have to travel a distance of d_r + l to make contact with the other side of the rocket (since both the ends of the chamber will have moved the same distance). So setting those equal:
d_b = d_r + l
Replacing d_b and d_r with the formulas and solving for l:
(v_0 t) = (v_0 t - \frac{1}{2} a_r t^2) + l
(v_0 t) - (v_0 t - \frac{1}{2} a_r t^2)= l
Collecting terms and simplifying:
(v_0 - v_0) t + \frac{1}{2} a_r t^2= l
(0) t + \frac{1}{2} a_r t^2= l
\frac{1}{2} a_r t^2= l
Solving for t:
t= \sqrt{\frac{2 l}{a_r}}.
Note that this doesn't depend on either the mass of the object or the velocity that it and the rocket are traveling at when the acceleration changes.