I. E. Irodov Solution 1.11

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: 1.11
Two particles move in a uniform gravitational field with an acceleration \(g\). At the initial moment the particles were located at one point and moved with velocities \({v_1} = 3m/s\) and \({v_2} = 4m/s\) horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors becomes mutually perpendicular.

Solution: 1.11
Here the two particle that was shot with \({v_1} = 3m/s\) and the one that was shot with \({v_2} = 4m/s\). Since \({v_1}\) and \({v_2}\) are supposed to be in opposite directions, we choose \({v_1}\) as the positive directions and \({v_2}\) as the negative direction. Further, we imagine that gravity acting downward is the negative direction.
Since, gravity acts vertically downwards, it does not effect the horizontal components of velocities. However both the particle accelerate at equal rates of \(g\) downwards from zero initial velocity component in downward direction.
The velocity of first particle after time \(t\) is given by \(\left( {{v_1}\hat i - gt\hat j} \right)\) and that second particle is given by \(\left( {{-
v_2}\hat i - gt\hat j} \right)\).
When the two velocity vectors are perpendicular to each other, the dot product of the vectors will be zero.
That is,
\(\left( {{v_1}\hat i - gt\hat j} \right).\left( { - {v_2}\hat i - gt\hat j} \right) = 0\)
Or, \(\left( { - {v_1}{v_2} + {g^2}{t^2}} \right) = 0\)
Or, \(\left( { - 12 + 100{t^2}} \right) = 0\)
Or, \(t = \sqrt {0.12} \)

The vertical distance traveled by the two particle will be identical, thus at any time \(t\) the distance between the two particles is only a function of the horizontal components of their positions is given by \(\left( {{v_1} + {v_2}} \right)t\).

Thus the distance between the two particles when their velocity vectors are mutually perpendicular are given by,
\( = \left( {{v_1} + {v_2}} \right) \times t = 7 \times \sqrt {0.12} = 2.5m\)