Question

TWO PARTICLES MOVE NEAR SURFACE OF EARTH WITH UNIFORM ACCELERATION g=10 TOWARDS GROUND . AT THE INITIAL MOMENT THE PARTICLES WERE LOCATED AT ONE POINT IN SPACE AND MOVED WITH VELOCITIES V1=3 AND V2=4 HORIZONTALLY IN OPPOSITE DIRECTION .FIND THE DISTANCE BETWEEN THE PARTICLES AT THE MOMENT WHEN THEIR VELOCITY VECTOR BECOMES MUTUALLY PERPENDICULAR.

Answer 1

Let us assume one particle is moving in direction of +ve x axis with $v_1$ velocity and the second particle in direction of -ve x axis with $v_2$ velocity.

Now the velocities of both the particles at time T:

When initial velocity of second particle is -4 and initial velocity of first particle is 3.

$v_2= (-4)i - (10t)j$

$v_1= (3)i - (10t)j$

When the vectors are at 90 degree angle with each other their product $v_{ 1 }\times v_{ 2 }$ gives 0

So

$0=-12+100{ t }^{ 2 }\quad \Rightarrow { t }^{ 2 }=\frac { 12 }{ 100 }$

Or $t=\frac { \left( \sqrt { 12 } \right) }{ 10 }$

At $t=\frac { \left( \sqrt { 12 } \right) }{ 10 }$, now the bodies have covered same distance that is the bodies will be in horizontal line

So the distance between the x coordinates of the bodies is the actual distance between the bodies

So the magnitudes are given as follows

$\frac { 3\left( \sqrt { 12 } \right) }{ 10 }\quad and\quad \frac { -4\left( \sqrt { 12 } \right) }{ 10 }$ are respectively the distance between particle A and B.

Or 1.03 and -1.38