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A hard physics question I'm stuck on

Hi,

I'm a little stuck on a physics question. Here it is:

We have two particles p1 and p2 in space. We know the velocity \(v_1\) and bearing \(\theta_1\) of p1, and we know the velocity \(v_2\) of p2. Find the bearing \(\theta_2\) that p2 should be so p1 and p2 will collide.

You can assume that \(p_1\) is on the point \((x_1,y_1)\) and \(p_2\) is at the origin.

I got a huge messy answer (courtesy of Wolfram Alpha for equation bashing):

\[\theta_2=2\tan^{-1}\left(\dfrac{\sqrt{2v_1^2x_1y_1\sin\theta_1\cos\theta_1+v_2^2(x_1^2+y_1^2)-v_1^2(x_1^2\cos^2\theta_1+y_1^2\sin^2\theta_1)}-v_2y_1}{v_2x_1+v_1(x_1\cos\theta_1-y_1\sin\theta_1)}\right)\]

And I am doubtful of its correctness.

A computer science approach is also welcome.

Any help is appreciated!

Note by Daniel Liu
1 year, 9 months ago

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A similar question is present in Irodov (1.5).

Two particles, \(1\) and \(2\), move with constant velocities \(\textbf{v}_1\) and \(\textbf{v}_2\). At the initial moment, their radius vectors are equal to \(\textbf{r}_1\) and \(\textbf{r}_2\). How must these four vectors be interrelated for the particles to collide?

Suppose the particles collide at time \(t_0\). Then their radius vectors must be equal at time \(t_0\). That is, \(\textbf{r}_1 + \textbf{v}_1 t_0= \textbf{r}_2 + \textbf{v}_2t_0\)

We get \(\textbf{r}_1 - \textbf{r}_2 = (\textbf{v}_2 - \textbf{v}_1)t_0 \). Since \(t_0\) is a positive constant, the vectors are parallel, and their unit vectors are equal. i.e.

\[ \frac{\textbf{r}_1 - \textbf{r}_2}{| \textbf{r}_1 - \textbf{r}_2 |} = \frac{\textbf{v}_2 - \textbf{v}_1}{|\textbf{v}_2 - \textbf{v}_1|}\]

We just need to use this interrelation to find \(\theta_2\).


I am converting the given information into vector form.

\[\textbf{r}_1 = \begin{pmatrix} x_1 \\y_1 \end{pmatrix} ~~~~~~~~~~~ \textbf{r}_2 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]

I am assuming the bearing to be taken counterclockwise from the positive \(x\) axis

\[\textbf{v}_1 = v_1 \begin{pmatrix} \cos \theta_1 \\ \sin \theta_1 \end{pmatrix} ~~~~~~~~~~~ \textbf{v}_2 = v_2 \begin{pmatrix} \cos \theta_2 \\ \sin \theta_2 \end{pmatrix}\]

After substituting in the derived relation, we get

\[\frac{1}{\sqrt{x_1^2 + y_1^2}} \begin{pmatrix} x_1 \\y_1 \end{pmatrix} = \frac{1}{\sqrt{(v_2 \cos \theta_2 - v_1 \cos \theta_1)^2 + (v_2 \sin \theta_2 - v_1 \sin \theta_ 1 )^2 }} \begin{pmatrix} v_2 \cos \theta_2 - v_1 \cos \theta_1 \\ v_2 \sin \theta_2 - v_1 \sin \theta_ 1 \end{pmatrix}\]

We obtain the following system of equations.

\[\begin{cases} x_1 \sqrt{(v_2 \cos \theta_2 - v_1 \cos \theta_1)^2 + (v_2 \sin \theta_2 - v_1 \sin \theta_ 1 )^2 )}= (v_2 \cos \theta_2 - v_1 \cos \theta_1)\sqrt{x_1 ^2 + y_1^2} \\ y_1 \sqrt{(v_2 \cos \theta_2 - v_1 \cos \theta_1)^2 + (v_2 \sin \theta_2 - v_1 \sin \theta_ 1 )^2 )} =(v_2 \sin \theta_2 - v_1 \sin \theta_1) \sqrt{x_1 ^2 + y_1^2} \end{cases}\]

Dividing the two equations, we get

\[\frac{x_1}{y_1} = \frac{v_2 \cos \theta_2 - v_1 \cos \theta_1}{v_2 \sin \theta_2 - v_1 \sin \theta_1} \]

\[x_1 (v_2 \sin \theta_2 - v_1 \sin \theta_1) = y_1 (v_2 \cos \theta_2 - v_1 \cos \theta_1)\]

\[ x_1 v_2 \sin \theta_2 - x_1 v_1 \sin \theta_1 = y_1 v_2 \cos \theta_2 - y_1 v_1 \cos \theta _1\]

\[v_2 (x_1 \sin \theta_2 - y_1 \cos \theta_2) = v_1 (x_1 \sin \theta_1 - y_1 \cos \theta _1)\]

Since this is of the form \(a \sin x - b\cos x = c\), we can use the R method.

\[\frac{x_1 \sin \theta_2 - y_1 \cos \theta_2}{\sqrt{x_1^2 + y_1^2}} = \frac{v_1}{v_2} \cdot \frac{(x_1 \sin \theta_1 - y_1 \cos \theta _1)}{\sqrt{x_1^2 + y_1 ^2}}\]

\[\sin (\theta_2 - \alpha) = \frac{v_1}{v_2} \cdot \frac{x_1 \sin \theta_1 - y_1 \cos \theta _1}{\sqrt{x_1^2 + y_1 ^2}}, \text{where } \tan \alpha = \frac{y_1}{x_1}\]

The R method can also be applied on the RHS. The equation simplifies to

\[v_1 \sin \left(\theta_1 - \alpha \right) = v_2 \sin \left(\theta_2 - \alpha \right)\]

\(\theta_2\) can be easily found out now.

I hope it helps. Pranshu Gaba · 1 year, 9 months ago

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@Pranshu Gaba Hello, bearing is defined to be the clockwise angle from North. However, thanks for your solution! I'll need to spend some time deciphering it.

I believe that the only thing that this difference in bearing definition changes is the definition \(\tan \alpha = \dfrac{x_1}{y_1}\), since it's just switching \(x\) and \(y\), is this correct?

EDIT: Ah, I actually also got the equation \(v_2(y_1\sin \theta_2-x_1\cos\theta_2) = v_1(y_1\sin\theta_1-x_1\cos\theta_1)\) from coordinate bashing and parameterizing. After this I didn't know what to do so I plugged it in Wolfram Alpha and got the above ugly expression. Thanks for teaching me (actually reminding me now that I think about it) the R method :) Daniel Liu · 1 year, 9 months ago

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@Daniel Liu Hi,

I am glad my solution reminded you of the R method!

Yes, the only difference the change in bearing definition makes is that \(\tan \alpha = \dfrac{x_1}{y_1}\) instead of \(\dfrac{y_1}{x_1}\). (After dividing the two equations, only \(x_1\) and \(y_1\) are interchanged - rest everything is same) While I knew that bearings are taken clockwise from North, I wasn't sure where the North pole is in the Cartesian plane.

I found it fascinating that the result is so neat! Do you think this problem can be generalised to \(n\) particles, or can we add more dimensions to the particles' position and velocity? The particles can also be made to accelerate! Pranshu Gaba · 1 year, 9 months ago

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@Pranshu Gaba @Pranshu Gaba Hello Pranshu. I reached the point till you said that the 2 vectors are parallel (but they could be coincident too, couldn't they?). Could you please explain how that makes their unit vectors equal, and how you got the next equation? Many thanks! Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra Hello Ishan, two vectors are equal when they have the same magnitude and direction.

We say that two vectors, say \(\vec{a}\) and \(\vec{b}\), are parallel, if there exists \(\lambda \neq 0\) such that \(\vec{a} = \lambda \vec{b}\). If \(\lambda\) is positive, then the two vectors have the same direction and if \(\lambda\) is negative, then the vectors point in opposite directions (In the above example, \(t_0\) is positive so the vectors are in the same direction).

When we divide both vectors by their magnitude, they both become unit vectors, and both their magnitudes become \(1\). Since the direction of both unit vectors is same and magnitude is also same, we can say that they are equal.

Note that translation does not change a vector. If we move the vector such that its magnitude and direction remain the same, then the vector does not change. If the vectors are coincident, translate one of them such that they become non-coincident, and then you can see that they are parallel; therefore coincident vectors are also parallel.

Can you tell me which next equation are you asking about?

Thanks. Pranshu Gaba · 1 year, 6 months ago

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@Pranshu Gaba @Pranshu Gaba \[\] Hello Pranshu. Unfortunately I've got into trouble with another Physics problem. Could you please show me where I've gone wrong?

Problem:

Two cars A and B move with velocity \(60 kmh^{-1}\) and \(70 kmh^{-1}\) from the same initial point. After a certain time, the two cars are \(2.5\) km apart. At that time, car B starts decelerating at the rate 20\(kmh^{-2}\). How long does Car A take to catch up with Car B?

\[\]I tried to apply Relative Motion Concept to try and solve this problem. However, I cannot understand how to apply it to this problem.

My attempt:\[\] I tried to apply the Relative Motion Concept to this problem as follows.

\[u_{AB}=u_A-u_B=60-70=-10kmh^{-1}\]

As per the question, the separation between the two cars ie \(S_{AB}=-2.5km\) after 15 minutes. \[\] Now, since Car A catches up with Car B eventually, thus \[S_{AB_{final}}=S_{A_{final}}-S_{B_{final}}=0\] and \[S_{AB_{initial}}=-2.5km \]\[\] \[\Longrightarrow \Large{S_{AB_{final}}-S_{AB_{initial}}=0-(-2.5)=u_{AB}\times t + \dfrac{1}{2}a_{AB}\times t^2}\]\[\] Now \[a_{AB}=a_A-a_B=0-(-20)=20kmh^{-2}\] and \[u_{AB}=-10kmh^{-1}\]\[\] \[\Longrightarrow 2.5=-10t+10t^2\]\[\] However, on solving this quadratic, I get a value of time which is incorrect. \[\]Could you please be so kind as to show me how to correctly apply the concept of Relative Motion here? I would be truly grateful for any assistance. Many thanks in advance! Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra @Josh Silverman Sir, please could you help me? Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra Sure, why don't you make a new note and mention me there Josh Silverman Staff · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra Hmm... Your solution appears to be correct, I can't find any error in it. Solving the quadratic equation gives time = 1.207 hours. What is the correct answer? Pranshu Gaba · 1 year, 6 months ago

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@Pranshu Gaba 0.5 hours. Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra @Nishant Rai Please could you help me? Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Ishan Dasgupta Samarendra The question is not specified clearly. To get an answer of \(0.5\), this is what the question probably means.

When the two cars are 2.5 km apart, car A is 2.5 km ahead of car B. So it is in fact car B which catches up to car A. \(S_{AB_{\text{initial}}}= + 2.5\)km . This gives the quadratic equation \(- 2.5 = -10 t + 10t^2\).

\(0.5\) is the root of this equation. Pranshu Gaba · 1 year, 6 months ago

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@Pranshu Gaba No, but the two cars start from the same initial point (I had forgotten to add that), and so \(S_{AB_{initial}}=-2.5 km\) Ishan Dasgupta Samarendra · 1 year, 6 months ago

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@Pranshu Gaba I just meant the one in which you divide the difference of 2 vectors by the modulus of the difference. I got it now- to make it a Unit Vector. Thanks a ton for your help! I'm really grateful to you. Ishan Dasgupta Samarendra · 1 year, 6 months ago

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Hint: Choose a suitable Initial Frame of reference. Calvin Lin Staff · 1 year, 9 months ago

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