I. E. Irodov Solution: 1.13

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE,...

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.

Problem: 1.13

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 \(u\). At the initial moment of time \(v\) perpendicular to \(u\) and the points are separated by a distance \(L\). How soon will the point converge?

Solution: 1.13

Figure shows an typical instantaneous state of the two particles during their motion as particle \(B\) is continuously aimed towards particle \(A\). Let \(x\) be the horizontal distance between the two particles and \(y\) be the vertical distance between the two particles. From second figure we can write the motion of the particle \(B\) given by the following equations,
\(\frac{{dx}}{{dt}} = u - v\cos \theta \) ... ... ... (i)
\(\frac{{dy}}{{dt}} = - v\sin \theta \) ... ... ... (ii)

Now, integrating both side (i) from \(t = 0\) to the time when \(A\) and \(B\) meet we obtain,
\(\int_0^0 {dx} = \int_0^\tau {v\cos \theta } - u\tau \)
Or, \(0 = \int_0^\tau {v\cos \theta } - u\tau \)
Or, \(\int_0^\tau {\cos \theta dt} = \frac{{u\tau }}{v}\) ... ... ... (iii)

The integral is from \(0\) to \(0\) in the above equation because the horizontal distance between \(A\) and \(B\) initially is \(0\) and finally when they meet also it will again \(0\). This can seen from the figure, from the definition of \(x\) as the horizontal distance between \(A\) and \(B\).

In the figure, we resolve all velocity along and perpendicular to the line connecting the two bodies.
If \(r\) is the distance between the two bodies then, as seen from the figure \(r\) decreases as \(\left( {v - u\cos \theta } \right)\) at any given instant of time.
Thus,
\(\frac{{dr}}{{dt}} = - \left( {v - u\cos \theta } \right)\)

The same result can of course also be derived by transforming to polar co-ordinates. Now if \(r\) is the vector connecting particles \(A\) and \(B\) then,
\(x = r\cos \theta \) and \(y = r\sin \theta \)
Or, \(\frac{{dx}}{{dt}} = \frac{{dr}}{{dt}}\cos \theta - r\sin \theta \frac{{d\theta }}{{dt}}\)
And, \(\frac{{dy}}{{dt}} = \frac{{dr}}{{dt}}\sin \theta + r\cos \theta \frac{{d\theta }}{{dt}}\)

Therefore,
\(\frac{{dr}}{{dt}}\cos \theta - r\sin \frac{{d\theta }}{{dt}} = u - v\cos \theta \)
And, \(\frac{{dr}}{{dt}}\sin \theta + r\cos \theta = - v\sin \theta \)

Or, \(\frac{{dr}}{{dt}}{\cos ^2}\theta - r\sin \theta \cos \theta = u\cos \theta - v{\cos ^2}\theta \)
And, \(\frac{{dr}}{{dt}}{\sin ^2}\theta + r\cos \theta \sin \theta \frac{{d\theta }}{{dt}} = - v{\sin ^2}\theta \)

Therefore,
\(\frac{{dr}}{{dt}} = u\cos \theta - v\)
Integrating both side, it would mean that,

\(\int_L^0 {dr} = \int_0^\tau {\left( {u\cos \theta - v} \right)dt} \)
Or, \(L = \int_0^\tau {\left( {v - u\cos \theta } \right)dt} \)
Or, \(L = v\tau - u\int_0^\tau {\cos \theta dt} \)

Now we can use the equation (ii) to solve for \(\tau \) as,
\(L = v\tau - u\left( {\frac{{u\tau }}{v}} \right)\)
Or, \(\tau = \frac{{Lv}}{{\left( {{v^2} - {u^2}} \right)}}\)