PHY.9:5.1 - Make mass affect the observation

Two bodies fall in a sufficiently uniform gravitational field with negligible drag. The equation m a=m g gives a=g. Timing their fall cannot distinguish their masses in this idealization, even though gravity exerts different forces.

Choose instead a known horizontal applied force F whose generation does not depend on the unknown mass. Track the body’s acceleration under that force. With other horizontal forces negligible, m a=F gives m=F/a for nonzero a. The physical change is from a force proportional to the unknown mass to an independently supplied force.

Now a constant horizontal force offset b matters during the short observation. Use two otherwise identical preparations with applied forces +F and -F. Assume the same b and the same mass in both, and measure their initial accelerations:

m a_+=F+b, m a_-=-F+b.

Subtracting gives

m=2F/(a_+-a_-).

With F=6 N, a_+=4 m/s² and a_-=-2 m/s², the mass is 2 kg and b=2 N. Using only F/a_+ would give 1.5 kg. The controlled reversal made mass distinguishable from the shared force offset.

The two preparations must preserve that offset. Velocity-dependent drag need not do so after the trajectories diverge. Measuring comparable initial responses, including the changed drag law, or selecting another arrangement can repair that use. Taking more samples of two physically different offsets does not justify the shared-b subtraction.

The acceleration readout and the force reference retain their own calibration and uncertainty. A decision whose margin exceeds those effects can use the mass estimate; a tighter use returns to the consequential contribution.