Kinematics

JEE Main level, single correct, medium

203 of 308

Question 203 of 308

An object is projected with kinetic energy KK from a point AA at an angle 60∘60^{\circ} with the horizontal. The ratio of the difference in kinetic energies at points BB and CC to that at point AA (see figure), in the absence of air friction is :

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Answer: (C) 3:43: 4

At A: KEA=KKE_A = K, vA=vv_A = v, angle =60°= 60°.
At B (highest point): vB=vcos⁡60°=v/2v_B = v\cos 60° = v/2, so KEB=K/4KE_B = K/4.
At C (same level as A): by energy conservation, KEC=KKE_C = K.
∣KEB−KEC∣=∣K/4−K∣=3K/4|KE_B - KE_C| = |K/4 - K| = 3K/4.
Ratio =3K/4K=34= \frac{3K/4}{K} = \frac{3}{4}, i.e., 3:43:4.

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