- #1
stunner5000pt
- 1,461
- 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: