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

Given:
Mass of ball, \( m = 100\,g = 0.1\,kg \) 
Velocity, \( v = 20\,m/s \)
Angle with horizontal = \( 60^\circ \)

Step 1: Initial Kinetic Energy

\[ K_i = \frac{1}{2} m v^2 \]

Step 2: Final Kinetic Energy

After the vertical component becomes zero, only the horizontal component \( v \cos 60^\circ \) remains. \[ K_f = \frac{1}{2} m (v \cos 60^\circ)^2 = \frac{1}{2} m \left( \frac{v}{2} \right)^2 = \frac{1}{8} m v^2 \]

Step 3: Decrease in Kinetic Energy

\[ \Delta K = K_i - K_f = \frac{1}{2} m v^2 - \frac{1}{8} m v^2 = \frac{3}{8} m v^2 \] Substitute values: \[ \Delta K = \frac{3}{8} \times 0.1 \times (20)^2 = \frac{3}{8} \times 0.1 \times 400 = 15\,J \]

Final Answer:

Decrease in kinetic energy = 15 J
✅ Correct Option: (2)

Answer 2

Step 1: Write given data.
\[ m = 100\,\text{g} = 0.1\,\text{kg}, \quad u = 20\,\text{m/s}, \quad \theta = 60^\circ. \]

Step 2: Separate velocity components.
\[ u_x = u \cos\theta = 20 \cos 60^\circ = 20 \times \frac{1}{2} = 10\,\text{m/s}, \] \[ u_y = u \sin\theta = 20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\,\text{m/s}. \]

Step 3: Kinetic energy at the point of projection.
\[ K_1 = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.1 \times (20)^2 = 0.05 \times 400 = 20\,\text{J}. \]

Step 4: Kinetic energy at the highest point.
At the highest point, the vertical velocity becomes zero (\( v_y = 0 \)), so only horizontal velocity remains: \[ v = u_x = 10\,\text{m/s}. \] \[ K_2 = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.1 \times (10)^2 = 0.05 \times 100 = 5\,\text{J}. \]

Step 5: Decrease in kinetic energy.
\[ \Delta K = K_1 - K_2 = 20 - 5 = 15\,\text{J}. \]

Wait! Let’s check: the question asks for “decrease in kinetic energy” — that is the difference \( K_1 - K_2 = 15\,\text{J} \). Hence the correct answer should be 15 J

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Final Answer:

\[ \boxed{\text{Decrease in kinetic energy} = 15\,\text{J}} \]