An object is projected with kinetic energy K from a point A at an angle 60° with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :

Question

G1 Physics · Accepted Answer

At point A, the initial kinetic energy is K = ½ μ \mu μ ². At the highest point B, the vertical velocity component is zero. The horizontal component remains u cos60° = u/2. Therefore, K(B) = ½m(u/2)² = ½m u²/4 = K/4. At point C, which is at the same vertical level as A, mechanical energy conservation gives K(C) = K. Hence the difference in kinetic energies at C and B is K(C) - K(B) = K - K/4 = 3K/4. The required ratio to the kinetic energy at A is (3K/4) : K = 3 : 4. Correct option: D.