Q3. A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²):
Explanation:
Velocity before hitting ground, v₁ = √(2×9.8×40) = 28 m/s
Velocity after bouncing up, v₂ = √(2×9.8×10) = 14 m/s
Impulse = m(v₂ - (-v₁)) = 0.5 × (14 + 28) = 21 Ns
Velocity before hitting ground, v₁ = √(2×9.8×40) = 28 m/s
Velocity after bouncing up, v₂ = √(2×9.8×10) = 14 m/s
Impulse = m(v₂ - (-v₁)) = 0.5 × (14 + 28) = 21 Ns
SEO Insight: This NEET question applies the impulse-momentum theorem and free fall motion. Practicing such questions boosts your understanding of how momentum changes during a bounce, a commonly tested concept in NEET Physics.
