Question

Two particles of masses 4 kg and 6 kg are at rest separated by 20 m. If they move towards each
other under mutual force of attraction, the position of the point where they meet is​

Answer 1

Answer:

12 m from 4 kg ball.

Explanation:

The mutual force produced by the particles is given by

F = GMm/r²

F = $\frac{3G}{25}$

acceleration of the first particle will be: F/M = $\frac{3G}{25}$ / 4

                                                                         = $\frac{3G}{100}$

                                                                          = 0.03 G

acceleration of second particle will be F/m = $\frac{3G}{25}$ / 6

                                                                        = $\frac{G}{50}$

                                                                         = 0.02 G

Say they meet at a point at a distance x from the 4 kg ball. (Have a look at the attachment below)

The time taken for the both the balls is the same.

So,

x = 1/2 * 0.03G * t²          (i) For 4kg ball

20-x = 1/2 * 0.02G * t²     (ii) For 6 kg ball

Dividing i and ii:

$\frac{x}{20-x}$ = $\frac{3}{2}$

2x = 60 - 3x

x = 12

Answer 2

Answer:

12m from 4kg block

Explanation:

hope this helps you mate ,......,