- #1

stunner5000pt

- 1,455

- 2

given a particle of mass m where r is the position vector for its trajectory

F is the force it experiences

p is its momentum

r0 represents the initial position vector

and r0 F, p are all constant vectors

show that

[tex] \vec{r}' = \frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t [/tex] where

[tex] \vec{r}' = \vec{r} - \vec{r_0} [/tex]

represents teh parametric equations of a parabola if F is not parallel to p

ok so the angular momentum of the particle is

[tex] \vec{L} = \vec{r}' \times \vec{p} = (\frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t) \times \vec{p} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) + \frac{t}{m} (\vec{p} \times \vec{p}) [/tex]

now certainly p cross p is zero

and we end up with

[tex] \vec{L} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) [/tex]

and this is not zero because F is not parallel to p

so this is a parabola because it is second order with responect to t?

this is the part that I am not too sure about...

Is this right at all first of all?

Thank you for your help!

F is the force it experiences

p is its momentum

r0 represents the initial position vector

and r0 F, p are all constant vectors

show that

[tex] \vec{r}' = \frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t [/tex] where

[tex] \vec{r}' = \vec{r} - \vec{r_0} [/tex]

represents teh parametric equations of a parabola if F is not parallel to p

ok so the angular momentum of the particle is

[tex] \vec{L} = \vec{r}' \times \vec{p} = (\frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t) \times \vec{p} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) + \frac{t}{m} (\vec{p} \times \vec{p}) [/tex]

now certainly p cross p is zero

and we end up with

[tex] \vec{L} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) [/tex]

and this is not zero because F is not parallel to p

so this is a parabola because it is second order with responect to t?

this is the part that I am not too sure about...

Is this right at all first of all?

Thank you for your help!

Last edited: