Two particle movement problem, apparent simple problem Two particles are given indexes '1' and '2'. They move in the xy plane with velocities $\vec{v_1}(t)=u\hat{p}$ and $\vec{v_2}(t)=v\hat{j}$, where $\hat{p}$ is the unit vector in the direction between the two particles: $\vec{p}(t)=\vec{r_2}(t)-\vec{r_1}(t)$, and $\hat{j}$ is the usual unitary vector in the y direction. $u,v$ are constants. At any moment $\vec{v_1},\vec{v_2}$ make an $\alpha(t)$ angle between each other.
a) Find a relation for $\dot{p}(t) \text{ and } \dot{\alpha}(t)$$$:
b)Show that they satisfy the relation: $$ \frac{dp}{d\alpha}=p\bigg(\frac{u-v\cos\alpha}{v\sin\alpha}\bigg)$$
Im having trouble with the second part. The first part I have show that:
$$\frac{d\vec{p}}{dt}=\vec{v_2}-\vec{v_1}=v\hat{j}-u\hat{p}\Rightarrow \frac{d\vec{p}}{dt}\cdot\frac{d\vec{p}}{dt}\equiv\big|{\dot{\vec{p}}}\big|^2=(v\hat{j}-u\hat{p})^2=v^2+u^2-2uv\hat{p}\cdot\hat{j}$$ thus $\dot{p}=\sqrt{v^2+u^2-2uv\hat{p}\cdot\hat{j}}$
The angle is such that: $$\vec{v_2}\cdot\vec{v_1}=cos\alpha|\vec{v_2}||\vec{v_1}|\Rightarrow cos\alpha = \hat{p}\cdot\hat{j}$$ thus $$\dot{p}=\sqrt{v^2+u^2-2uvcos\alpha} $$
Finding $\alpha$ gets hard as I go and the b) item is further more. Someone might have a lead on this?
 A: Cool task!
I think your expression for $\dot p$ is slightly off. Notice first that
$$ \vec{p} \cdot \dot{\vec{p}} = p \hat{p} \cdot \dot{\vec{p}} = p \hat{p} \cdot \left( v \hat{j} - u \hat{p} \right) = pv \cos(\alpha) - pu
$$
The quantity $p=p(t)$ is the length of the vector $\vec{p}$, and the quantity $\dot{p}$ is the time derivative of the length of $\vec{p}$, so
$$ \dot{p} = \frac{d}{dt} ｜\vec{p} ｜ = \frac{d}{dt} \sqrt{\vec{p} \cdot \vec{p} } = \frac{1}{2 \sqrt{\vec{p} \cdot \vec{p} } } \frac{d}{dt}(\vec{p} \cdot \vec{p}) = \frac{1}{2p}2\vec{p} \cdot \dot{\vec{p}}
$$
where the product rule has been used in the last step. Based on what we found above, this gives
$$ \dot{p} = \frac{1}{p}(pv \cos(\alpha) - pu) = v \cos(\alpha) - u
$$
which is not identical to your expression.
Starting from the expression
$$ \cos(\alpha) = \hat{p} \cdot \hat{j}
$$
we quickly find that
$$ -\sin(\alpha) \dot{\alpha} = \dot{\hat{p}} \cdot \hat{j} 
$$
Further, we note that we can write $\dot{\vec{p}}$ as
$$ \dot{\vec{p}} = \dot{p} \hat{p} + p \dot{\hat{p}}
$$
using the product rule, and if we dot this with $\hat{j}$ we find
$$(v \hat{j} - u \hat{p}) \cdot\hat{j} = \dot{p} \hat{p} \cdot \hat{j} + p \dot{\hat{p}} \cdot \hat{j}
$$
$$v - u \cos(\alpha) = \dot{p} \cos(\alpha) -p \sin(\alpha) \dot{\alpha}
$$
Inserting our expression for $\dot{p}$ from before and doing some algebra results in
$$ p \dot{\alpha} = - v \sin( \alpha)
$$
Now we can use the chain rule
$$ \dot{p} = \frac{dp}{d\alpha} \dot{\alpha},
$$
multiply both sides by $p$ and again use our expression for $\dot{p}$ to get
$$p(v \cos(\alpha) - u) = \frac{dp}{d\alpha} (- v \sin( \alpha))
$$
The desired result for b) follows.
