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 at points B and C to that at point A (see figure), in the absence of air friction is : [figure]
- 3 : 4
- 1 : 4
- 2 : 3
- 1 : 2
Answer
(1) 3 : 4
Solution
Let the speed of projection be u, so K = 12mu2.
At the highest point B only the horizontal component remains: vB = u 60 = u2
KB = 12m ( u2 )2 = K4
C is on the same horizontal level as A, so by conservation of mechanical energy KC = KA = K
KC - KB = K - K4 = 3K4
Required ratio = KC - KBKA = 3K/4K = 34, i.e. 3 : 4
Answer: (1)