To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as

As M=m, then

Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then



Therefore replacing we have that,

Re-arrange to find v,



Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that

Therefore




Therefore the velocity when they are about to collide is 
Similarity : inverse square law for strength of force compared with distance.
I = MR^2
The Attempt at a Solution:::
I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2
I total = 3ML^2/2
It says the answer is 3ML^2/4 though.
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Answer:
The kinetic energy of the particle as it moves through point B is 7.9 J.
Explanation:
The kinetic energy of the particle is:
<u>Where</u>:
K: is the kinetic energy
: is the potential energy
q: is the particle's charge = 0.8 mC
ΔV: is the electric potential = 1.5 kV
Now, the kinetic energy of the particle as it moves through point B is:


Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.
I hope it helps you!
Answer:
in everyday use and in kinematic the speed of an object is the magnitude of the change of its position it is thus a scalar quantity