June 13, 2008

Q. 1.11 - "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$:

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 $v_1$= 3.0 m/s and $v_2$= 4.0 m/s horizontally in opposite directions. find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

$\textbf{Answer}$:

Since there is no horizontal force acting on both the particles, the horizontal component of their velocity remains same.

Representing the particle velocities as vectors,

Velocity vector of the first particle at time $t$= $v_1\vec{i} + gt\vec{j}$(assuming $\vec{j}$as the direction in which the gravitational force is acting).

Velocity vector of the second particle at time $t$= $-v_2\vec{i} + gt \vec{j}$.

Angle between these two velocity vectors at time $t$=

\begin{displaymath}\tan^{-1}\frac{gt}{v_1} - \tan^{-1}\frac{gt}{-v_2}\end{displaymath}

\begin{displaymath}= \tan^{-1}\frac{gt(v_1 + v_2)}{g^2 t^2 - v_1 v_2} \end{displaymath}

When this angle is $\pi/2$, $g^2 t^2 = v_1 v_2$. Therefore, $t = \frac{1}{g}\sqrt{v_1 v_2} $.

And at this time, the distance between the two particles is $v_1 t + v_2 t$= $\frac{(v_1 + v_2)\sqrt{v_1 v_2}}{g}$

Using values, we get the distance = $\frac{7 \times \sqrt{12}}{9.8}$= 2.47m

Posted by Apurva Sharan at 10:09 AM
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