Question

We initially place a particle of mass 1 kg at \(\vec r = \hat i \) and give it an initial velocity \(\vec v = \hat j\). A force starts acting on the particle and is given by \(\vec F = - \vec r ~N\). What is the trajectory of the particle?

Answer 1

\[\vec F = m \dfrac{d\vec v}{dt} = -\vec r\]\[\Rightarrow d \vec v = -\frac{1}{m}\vec r dt\]

Answer 2

Hmm, maybe apply the relation\[\vec a = v \dfrac{dv}{dr}\]

Answer 3

\[-\vec r/m ~dr= v ~dv\]Yeah, that works.

Answer 4

-r dr = v dv

r^2/2 from 1i to r = v^2/2 from 1j to v.

Answer 5

Does it mean anything to square a vector? Should it be more like v.v or r.r?

Answer 6

...which is where I'm stuck.

Answer 7

Oh yeah, we can square a vector like that.

Answer 8

Well yeah, but that's very qualitative. We'd need an equation to actually verify our work.

Answer 9

is that F=<1,0> then
or is that saying F is directioly proportional to the vector position of the pariticle R

Answer 10

F = <-x, -y>

dan

Answer 11

The force acts opposite to the position vector at all points.

Answer 12

kk so r(0)=<1,0> basically

Answer 13

Oh, so do I work with the component form?