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. Zero
• B. 15 J
• C. 20 J
• D. 5 J

Hint:

The decrease in kinetic energy in projectile motion corresponds to the loss of vertical velocity as the object reaches the highest point. The horizontal component remains unchanged.

Answer 1

The Correct Option is B

Provided:
Mass of the ball, \( m = 100\,g = 0.1\,kg \)
Initial velocity, \( v = 20\,m/s \)
Launch angle relative to the horizontal = \( 60^\circ \)

Step 1: Calculate Initial Kinetic Energy

The initial kinetic energy is given by: \[ K_i = \frac{1}{2} m v^2 \]

Step 2: Calculate Final Kinetic Energy

Upon reaching the apex of its trajectory, the vertical component of velocity becomes zero. The remaining horizontal velocity is \( v \cos 60^\circ \). The final kinetic energy is: \[ K_f = \frac{1}{2} m (v \cos 60^\circ)^2 = \frac{1}{2} m \left( \frac{v}{2} \right)^2 = \frac{1}{8} m v^2 \]

Step 3: Determine the Decrease in Kinetic Energy

The change in kinetic energy is the difference between the initial and final kinetic energies: \[ \Delta K = K_i - K_f = \frac{1}{2} m v^2 - \frac{1}{8} m v^2 = \frac{3}{8} m v^2 \] Substituting the given values: \[ \Delta K = \frac{3}{8} \times 0.1 \times (20)^2 = \frac{3}{8} \times 0.1 \times 400 = 15\,J \]

Final Result:

The kinetic energy decreased by 15 J
✅ Corresponding Option: (2)