Question

A ball of mass 100 g is projected with velocity 20 m/s at $60^\circ$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is:

• A. 5 J
• B. 15 J
• C. 20 J
• D. zero

Hint:

In projectile motion, the loss in kinetic energy up to the highest point depends only on the vertical component of velocity.

Answer 1

The Correct Option is B

To determine the decrease in kinetic energy of the ball during its motion from the point of projection to the highest point, we need to analyze the motion of the ball considering its initial and final velocity components.

1. The ball is projected with an initial velocity \( u = 20 \, \text{m/s} \) at an angle \( \theta = 60^\circ \) to the horizontal.
2. We first resolve this initial velocity into its horizontal and vertical components:
   • Horizontal component of velocity, \( u_x = u \cos \theta = 20 \cos 60^\circ = 20 \times 0.5 = 10 \, \text{m/s} \)
   • Vertical component of velocity, \( u_y = u \sin \theta = 20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 20 \times 0.866 \approx 17.32 \, \text{m/s} \)
3. At the highest point of its trajectory, the vertical component of the velocity becomes zero (as the ball momentarily stops moving upward before descending): \( v_y = 0 \).
4. The horizontal component of velocity remains unchanged throughout the projectile motion, as there is no acceleration in the horizontal direction: \( v_x = u_x = 10 \, \text{m/s} \).
5. Calculate the initial kinetic energy at the point of projection:
   • Total initial kinetic energy \( K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (20)^2 \, \text{m}^2/\text{s}^2 = \frac{1}{2} \times 0.1 \times 400 = 20 \, \text{J} \)
6. Calculate the kinetic energy at the highest point, considering only the horizontal component of velocity:
   • Total kinetic energy at highest point \( K_h = \frac{1}{2} m v_x^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (10)^2 \, \text{m}^2/\text{s}^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \)
7. The decrease in kinetic energy is the difference between the initial kinetic energy and the kinetic energy at the highest point:
   • Decrease in kinetic energy \( = K_i - K_h = 20 \, \text{J} - 5 \, \text{J} = 15 \, \text{J} \)

Therefore, the decrease in kinetic energy of the ball during the motion from the point of projection to the highest point is 15 J.