A bob of mass ' m ' is suspended by a light string of length ' L '. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies ( K.E. )A( K.E. )B is :

The problem involves the conservation of mechanical energy, since no non-conservative forces are doing work.

At the lowest point A, the bob has maximum kinetic energy and potential energy is zero (taking potential energy as zero at the lowest point). Let's denote the minimum horizontal velocity needed to reach the topmost point B as v. So, the total energy at A is:

E(A) = K.E.(A) + P.E.(A) = 1/2 * m * v^2 + 0 = 1/2 * m * v^2

At the topmost point B, the bob has minimum kinetic energy and maximum potential energy. The height h of point B from point A is equal to the diameter of the circle, which is 2L. So, the total energy at B is:

E(B) = K.E.(B) + P.E.(B) = 1/2 * m * v1^2 + m * g * 2L

Since mechanical energy is conserved, E(A) = E(B). Therefore, we have:

1/2 * m * v^2 = 1/2 * m * v1^2 + m * g * 2L

Solving for v1^2, we get:

v1^2 = v^2 - 4g * L

The ratio of kinetic energies at A and B is then:

(K.E.(A))/(K.E.(B)) = (1/2 * m * v^2) / (1/2 * m * v1^2) = v^2 / (v^2 - 4g * L)

This is the ratio of kinetic energies at the lowest point A and the topmost point B.