Question

An object is projected with kinetic energy K from 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 : 

• A. 2 : 3
• B. 1 : 2
• C. 3 : 4
• D. 1 : 4

Hint:

At the maximum height of a projectile, the kinetic energy is never zero; it is equal to $K \cos^2 \theta$.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the ratio of the difference in kinetic energies at points B and C to that at point A for a projectile motion, given that the object is projected from point A with kinetic energy \( K \) at an angle of \( 60^\circ \) with the horizontal. We'll assume there is no air friction.

Let's break down the problem step by step:

1. Initially, at point A, the object is projected with a kinetic energy \( K \).
2. At point B, the object is at its maximum height, so its vertical component of velocity is zero. Therefore, the kinetic energy at B, \( K_B \), is due to the horizontal component only.
3. The horizontal component of velocity \( v_{x} \) remains constant throughout the motion because there's no air friction. Let's denote the initial velocity as \( v_0 \).
4. At point A, the velocity components are:
   • Horizontal component: \( v_{0x} = v_0 \cos(60^\circ) = \frac{v_0}{2} \)
   • Vertical component: \( v_{0y} = v_0 \sin(60^\circ) = \frac{\sqrt{3}v_0}{2} \)
5. The initial kinetic energy is given by: \(K = \frac{1}{2}m v_0^2\)
6. At point B (at maximum height): \(K_B = \frac{1}{2} m \left(\frac{v_0}{2}\right)^2 = \frac{1}{8}m v_0^2 = \frac{K}{4}\)
7. At point C, the object has returned to the initial vertical level as point A, so the kinetic energy \( K_C \) is equal to the initial kinetic energy: \(K_C = K\)
8. The difference in kinetic energies at points B and C compared to point A is:
   • At B: \( \Delta K_B = K - K_B = K - \frac{K}{4} = \frac{3K}{4} \)
   • At C: \( \Delta K_C = K - K = 0 \)
9. So the ratio of the difference in kinetic energies at points B and C to that at point A is: \(\frac{K - K_B}{K} = \frac{\frac{3K}{4}}{K} = \frac{3}{4}\)

Therefore, the correct answer is 3 : 4.