A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g = 10 m s-2)

To find the net force, we use the Impulse-Momentum Theorem, which states that the net force is equal to the rate of change of momentum (F = Δp / Δt).

• Step 1: Calculate initial velocity (v₁) just before hitting the floor: v₁ = √(2gh₁) = √(2 × 10 × 0.45) = √9 = 3 m/s (downward).
• Step 2: Calculate final velocity (v₂) just after bouncing: v₂ = √(2gh₂) = √(2 × 10 × 0.20) = √4 = 2 m/s (upward).
• Step 3: Calculate change in momentum (Δp): Since direction changes, Δp = m(v₂ - (-v₁)) = m(v₂ + v₁) = 0.1 × (2 + 3) = 0.5 kg·m/s.
• Step 4: Calculate Net Force (F): F = Δp / Δt = 0.5 / 0.1 = 5.0 N.

The net force applied during the collision, derived from the change in momentum over the contact time, is 5.0 N, aligning with Option D.