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Problem 1.13 from Irodov's Problems in General Physics:
Point A moves uniformly with velocity $v$ so that the vector $v$ is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity $u<v$. Initially, $v⊥u$, and the points are separated by a distance $l$. The time at which the points converge is given as?
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If we write $\tau$ as the time to collision, then we can immediately find $$\int_0^\tau v\cos\theta\,dt=u\tau\Leftrightarrow\int_0^\tau\cos\theta\,dt=\frac{u\tau}{v}\\\int_0^\tau v\sin\theta\,dt=l\Leftrightarrow\int_0^\tau\sin\theta\,dt=\frac{l}{v}$$ Irodov asserts that $$\int_0^\tau(v-u\cos\theta)dt=l$$ It turns out that if we consider $\vec w=\vec v-\vec u=(v\cos\theta-u,v\sin\theta)$, then $v-u\cos\theta$ is $v\cos\theta-u$ in the basis $(\hat v,\hat v_\perp)$, where $\hat v=(\cos\theta, \sin\theta)$ and $\hat v_\perp=(\sin\theta,-\cos\theta)$ in $(\hat\imath,\hat\jmath)$.
$\vec w=(v-u\cos\theta,-u\sin\theta)$ in $(\hat v,\hat v_\perp)$.

Consider integrating $\vec w$ (I use the integrals of the sine and cosine above in the process) $$\int_0^\tau\vec w\,dt=\left(\int_0^\tau(v\cos\theta-u)\hat\imath dt\right)+\left(\int_0^\tau(v\sin\theta)\hat\jmath dt\right)=l\hat\jmath$$ Now consider the same but with the coordinates in $(\hat v,\hat v_\perp)$ $$\int_0^\tau\vec w\,dt=\left(\int_0^\tau(v-u\cos\theta)\hat v dt\right)+\left(\int_0^\tau(-u\sin\theta)\hat v_\perp dt\right)=\left(\frac{v^2-u^2}{v}\tau\right)\hat v-\left(\frac{ul}{v}\right)\hat v_\perp$$ If we equate both integrals and take the norm squared, we get $$\left(\frac{v^2-u^2}{v}\tau\right)^2+\left(\frac{ul}{v}\right)^2=l^2\Leftrightarrow\tau=\frac{l}{\sqrt{v^2-u^2}}$$ But this is wrong, the actual value is $$\tau=\frac{vl}{v^2-u^2}$$ Can anyone spot the error? Thank you!

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  • $\begingroup$ $\hat v$ and $\hat v_\perp$ are functions of $t$. Here is a different formulation of the book's approach. $\endgroup$
    – Maxim
    Jan 13, 2021 at 17:44
  • $\begingroup$ @Maxim thank you for your answer. I have guessed as much, but I don't see why that would cause a problem since we now have those as fixed vectors. Remember that I have done a transition from cartesian coordinates to $(\hat v,\hat v_\perp)$. $\endgroup$
    – GDGDJKJ
    Jan 18, 2021 at 4:32
  • $\begingroup$ In the reference frame $\mathcal R$ in which the problem is stated, integrating $x \hat i + y \hat j = x' \hat v + y' \hat v_\perp$ wrt $t$ cannot give a result that still depends on the components of $\hat v$ and $\hat v_\perp$ in $\mathcal R$ and therefore on $t$. If you wish to switch to a different reference frame $\mathcal R'$, you have to specify not just the basis vectors but also how the origin of $\mathcal R'$ moves wrt the origin of $\mathcal R$. $\endgroup$
    – Maxim
    Jan 22, 2021 at 16:45

2 Answers 2

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Let $A(t),B(t)$ denote the positions of the bodies at time $t\geq 0$. Take $B(0)=(0,0)$ and $A(0)=(0,-l)$. The information provided in the problem tells us $\frac{dB}{dt}=\Big<u,0\Big>$ and $$\frac{dA}{dt}=\frac{v\big(B(t)-A(t)\big)}{||B(t)-A(t)||}$$ If we take $\vec{p}(t)=B(t)-A(t)$ we can say $$\frac{d\vec{p}}{dt}=\Big<u,0\Big>-v\frac{\vec{p}(t)}{||\vec{p}(t)||}$$ We need to find $\tau$ so that $\vec{p}(\tau)=\vec{0}$ Writing $\vec{p}(t)=\big(x(t),y(t)\big)$ gives us the system $$\frac{dx}{dt}=u-\frac{vx}{\sqrt{x^2+y^2}}$$ $$\frac{dy}{dt}=-\frac{vy}{\sqrt{x^2+y^2}}$$ Converting the polar yields $$\frac{dr}{dt}=u\cos(\theta)-v$$ $$\frac{d\theta}{dt}=-\frac{u\sin(\theta)}{r}$$ Here $(r,\theta)=(l,\pi/2)$ when $t=0$. Dividing these provides us with a separable DE in $r$ and $\theta$: $$\frac{dr}{d\theta}=r\cdot \Bigg(\frac{v-u\cos(\theta)}{u\sin(\theta)}\Bigg)$$ This has solution $$r=\frac{l\big(\sin(\theta)\big)^{\frac{v}{u}-1}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}$$ Where $\theta \in [0,\pi/2]$ and $r=||\vec{p}(t)||=0$ when $\theta=0$. Plugging into our formula for $\frac{d\theta}{dt}$ gives us $$\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=-\frac{u}{l}dt$$ This means $$\int_{\pi/2}^0\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta=\int_{0}^{\tau}-\frac{u}{l}dt$$ Solving for $\tau$, $$\tau=\frac{l}{u}\int_{0}^{\pi/2}\Bigg(\frac{\big(\sin(\theta)\big)^{\frac{v}{u}-2}}{\big(\cos(\theta)+1\big)^{\frac{v}{u}}}\Bigg)d\theta$$ I know this isn't the answer you're looking for but it works!

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Given that A always aims at B, establish the equation below $$\frac{v_y}{v_x}=\frac{-y}{ut-x}$$

Rewrite it as $x-ut = y\frac{v_x}{v_y}$ and differentiate both sides wrt. time $$-\frac uy = \frac{d}{dt} \left(\frac{v_x}{v_y}\right)\implies \frac uv \frac {dy}y = \frac{v\>{dv_y}}{v_y^2\sqrt{\frac{v^2}{v_y^2}-1}} $$ where $v^2=v_x^2+v_y^2$ and $\frac{dv_y}{dt}=v_y \frac{dv_y}{dy}$ are applied. Integrate both sides to arrive at

$$\frac uv \ln\left(-\frac{y}l\right) = \cosh^{-1}\frac v{v_y} $$

Then, the collision time can be obtained as follows

$$\tau = \int_0^\tau dt = \int_{-l}^0 \frac{dy}{v_y} =\frac1{v} \int_{-l}^0 \cosh\left[\frac uv \ln \left(-\frac{y}l\right) \right] dy=\frac{vl}{v^2-u^2} $$

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