Question

Initially two particles A and B are present at (0,
and (d, 0) respectively. They start moving with
VA=vi + vj
velocity
and
B
is magnitude of relative separation between th
and be the time when separation between
them is a minimum, then
Vo = -vj.
To
clear respon​

Answer 1

Explanation:

we know,

C = klo A upon d

Now distance is halved K id doubled

C1 = (2k) 10A upon ( d upon 2)

C1 = 4C

Answer 2

The complete question is:

Initially two particles A and B are present at (0,0) and (d,0) respectively They start moving with speed VA=Viˆ+Vjˆ and VB=−Vjˆ .

If R is magnitude of relative separation between them and T0 be the time when separation between them is minimum, then

1. T0=d/5V
2. Rmin=2d/√5
3. Graph of R versus time is straight line
4. Graph of R versus time is circle.

Given:

Initial position of A = (0,0)

Initial position of B = (d,0)

Initial velocity of A = Viˆ + Vjˆ

Initial velocity of B = -Vj

To find:

The correct statement from the given options.

Solution:

The figure is attached here.

If we take B as frame of reference :

Relative velocity of A with respect to B is;

vAB=vA−vB=viˆ+2vjˆ

Magnitude of velocity of A= (v^2 + v^2)^1/2

= √2 v

Magnitude of velocity of B = v

Magnitude of velocity of A with respect to B = (v²+ 4v²)^1/2 = √5 v

And the angle between the vectors of vAB is:

α=tan−1(2)

According to the figure.

rmin=dsinα=2d/√5

T0= AM/∣vA/B∣ = d/√5

R2=(5–√vtcosα−d)^2 + (5–√vtsinα)^2

Since the equation neither resembles the equation of circle or straight line therefore the graph is neither circle nor straight line      

The correct option is option 1 and 2.