Question

Two equal masses of 6.40 kg are separated by a distance of 0.16m a small body is releases from the point p equidistant from the two masses and at a distance of 0.06m from the line joining them . 1)calculate the velocity of the body when it passes through the q 2)calculate the acceleration of this body at p and q if it's mass is 0.1kg

Answer 1

At point P
The force would be
F = 6.67*10^-11 * 6.4 *0.1/0.1^2
F = 426.8 * 10^-11
As their is 2 bodies of same mass the force by the second body would be the same and angle between those 2 forces is 90° and hence the resultant force would be of
F(r) = (426.8*10^11)^2 + (426.8*10^11)^2
We get is
3.64316 * 10^-27 N
And hence dividing by the total mass we get is
0.2824155*10^-27 m/sec^2
And now at point Q
The acceleration would be ZERO
AS those 2 forces oppose each other
Now the velocity would be we have
Displacement would be 0.06 m now
S = ut+1/2at^2
0.06*2 = 0.2824155 t^2
0.4249 = t^2
T = 0.64 sec
Now acc = change in velocity/time
0.2824155 = v(f)/time
0.2824155*0.65 = Velocity
But 1 thing after this value of time even a microsecond later the velocity will become zero as at this point the acceleration now become zero thus stopping the particle to move further