A hard physics question I'm stuck on


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 v1v_1 and bearing θ1\theta_1 of p1, and we know the velocity v2v_2 of p2. Find the bearing θ2\theta_2 that p2 should be so p1 and p2 will collide.

You can assume that p1p_1 is on the point (x1,y1)(x_1,y_1) and p2p_2 is at the origin.

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


And I am doubtful of its correctness.

A computer science approach is also welcome.

Any help is appreciated!

Note by Daniel Liu
6 years, 3 months ago

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A similar question is present in Irodov (1.5).

Two particles, 11 and 22, move with constant velocities v1\textbf{v}_1 and v2\textbf{v}_2. At the initial moment, their radius vectors are equal to r1\textbf{r}_1 and r2\textbf{r}_2. How must these four vectors be interrelated for the particles to collide?

Suppose the particles collide at time t0t_0. Then their radius vectors must be equal at time t0t_0. That is, r1+v1t0=r2+v2t0\textbf{r}_1 + \textbf{v}_1 t_0= \textbf{r}_2 + \textbf{v}_2t_0

We get r1r2=(v2v1)t0\textbf{r}_1 - \textbf{r}_2 = (\textbf{v}_2 - \textbf{v}_1)t_0 . Since t0t_0 is a positive constant, the vectors are parallel, and their unit vectors are equal. i.e.

r1r2r1r2=v2v1v2v1 \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 θ2\theta_2.

I am converting the given information into vector form.

r1=(x1y1)           r2=(00)\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 xx axis

v1=v1(cosθ1sinθ1)           v2=v2(cosθ2sinθ2)\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

1x12+y12(x1y1)=1(v2cosθ2v1cosθ1)2+(v2sinθ2v1sinθ1)2(v2cosθ2v1cosθ1v2sinθ2v1sinθ1)\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.

{x1(v2cosθ2v1cosθ1)2+(v2sinθ2v1sinθ1)2)=(v2cosθ2v1cosθ1)x12+y12y1(v2cosθ2v1cosθ1)2+(v2sinθ2v1sinθ1)2)=(v2sinθ2v1sinθ1)x12+y12\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

x1y1=v2cosθ2v1cosθ1v2sinθ2v1sinθ1\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}

x1(v2sinθ2v1sinθ1)=y1(v2cosθ2v1cosθ1)x_1 (v_2 \sin \theta_2 - v_1 \sin \theta_1) = y_1 (v_2 \cos \theta_2 - v_1 \cos \theta_1)

x1v2sinθ2x1v1sinθ1=y1v2cosθ2y1v1cosθ1 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

v2(x1sinθ2y1cosθ2)=v1(x1sinθ1y1cosθ1)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 asinxbcosx=ca \sin x - b\cos x = c, we can use the R method.

x1sinθ2y1cosθ2x12+y12=v1v2(x1sinθ1y1cosθ1)x12+y12\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(θ2α)=v1v2x1sinθ1y1cosθ1x12+y12,where tanα=y1x1\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

v1sin(θ1α)=v2sin(θ2α)v_1 \sin \left(\theta_1 - \alpha \right) = v_2 \sin \left(\theta_2 - \alpha \right)

θ2\theta_2 can be easily found out now.

I hope it helps.

Pranshu Gaba - 6 years, 3 months ago

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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 tanα=x1y1\tan \alpha = \dfrac{x_1}{y_1}, since it's just switching xx and yy, is this correct?

EDIT: Ah, I actually also got the equation v2(y1sinθ2x1cosθ2)=v1(y1sinθ1x1cosθ1)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 - 6 years, 3 months ago

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

Yes, the only difference the change in bearing definition makes is that tanα=x1y1\tan \alpha = \dfrac{x_1}{y_1} instead of y1x1\dfrac{y_1}{x_1}. (After dividing the two equations, only x1x_1 and y1y_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 nn particles, or can we add more dimensions to the particles' position and velocity? The particles can also be made to accelerate!

Pranshu Gaba - 6 years, 3 months ago

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@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!

User 123 - 6 years, 1 month ago

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

We say that two vectors, say a\vec{a} and b\vec{b}, are parallel, if there exists λ0\lambda \neq 0 such that a=λb\vec{a} = \lambda \vec{b}. If λ\lambda is positive, then the two vectors have the same direction and if λ\lambda is negative, then the vectors point in opposite directions (In the above example, t0t_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 11. 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?


Pranshu Gaba - 6 years, 1 month 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.

User 123 - 6 years 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?


Two cars A and B move with velocity 60kmh160 kmh^{-1} and 70kmh170 kmh^{-1} from the same initial point. After a certain time, the two cars are 2.52.5 km apart. At that time, car B starts decelerating at the rate 20kmh2kmh^{-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.


As per the question, the separation between the two cars ie SAB=2.5kmS_{AB}=-2.5km after 15 minutes. Now, since Car A catches up with Car B eventually, thus SABfinal=SAfinalSBfinal=0S_{AB_{final}}=S_{A_{final}}-S_{B_{final}}=0 and SABinitial=2.5kmS_{AB_{initial}}=-2.5km SABfinalSABinitial=0(2.5)=uAB×t+12aAB×t2\Longrightarrow \Large{S_{AB_{final}}-S_{AB_{initial}}=0-(-2.5)=u_{AB}\times t + \dfrac{1}{2}a_{AB}\times t^2} Now aAB=aAaB=0(20)=20kmh2a_{AB}=a_A-a_B=0-(-20)=20kmh^{-2} and uAB=10kmh1u_{AB}=-10kmh^{-1} 2.5=10t+10t2\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!

User 123 - 6 years ago

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@User 123 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 - 6 years ago

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@Pranshu Gaba 0.5 hours.

User 123 - 6 years ago

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@User 123 The question is not specified clearly. To get an answer of 0.50.5, 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. SABinitial=+2.5S_{AB_{\text{initial}}}= + 2.5km . This gives the quadratic equation 2.5=10t+10t2- 2.5 = -10 t + 10t^2.

0.50.5 is the root of this equation.

Pranshu Gaba - 6 years 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 SABinitial=2.5kmS_{AB_{initial}}=-2.5 km

User 123 - 6 years ago

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@User 123 @Nishant Rai Please could you help me?

User 123 - 6 years ago

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@User 123 @Josh Silverman Sir, please could you help me?

User 123 - 6 years ago

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@User 123 Sure, why don't you make a new note and mention me there

Josh Silverman Staff - 6 years ago

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Hint: Choose a suitable Initial Frame of reference.

Calvin Lin Staff - 6 years, 3 months ago

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