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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 im 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 im not too sure about...

Is this right at all first of all?

Thank you for your help!

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