# What does it mean for a particle to be free?

## Homework Statement

From an inertial reference frame S, the vector position of a particle of mass
m1 = 1kg is given by r1(t)=(tx-hat - t^2y-hat)m.
The vector position of a particle m2=2m1 is given by r2=(t)=(tx-hat +t^3y-hat)m

Is particle #1 free along the y-direction? Explain

## Homework Equations

If the particle is free along the y-direction then the y's would equal to 0.

## The Attempt at a Solution

r1= -t^2y
r2=+t^3y
=not free?
(I am actually unsure if my definition of free is true)

Fredrik
Staff Emeritus
Gold Member
"Free" means "unaffected by forces", or equivalently that the potential is constant. In this case, it means you have to check if it's accelerating in the y-direction or not.

Particle #1's motion along the y-direction is -t^2.
Would this mean that it is not free?

Fredrik
Staff Emeritus