Two ball of same mass are dropped from the same height onto the floor. The first ball bounces upwards from the floor elastically. The second ball sticks to the floor. The first applies an impulse to the floor of I1 and the second applies an impulse I2. The impulses obey

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

Two ball of same mass are dropped from the same height onto the floor. The first ball bounces upwards from the floor elastically. The second ball sticks to the floor. The first applies an impulse to the floor of I₁ and the second applies an impulse I₂. The impulses obey

I₂=2I₁

I₂=I₁/2

I₂=4I₁

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Impulse (I) equals change in momentum (Δp). Both balls have same mass (m) and initial downward velocity (v) before impact.

For elastic bounce: velocity reverses direction. Change in momentum: ${Δp}_{1} = m (v - (- v)) = 2 m v$ . So $I_{1} = 2 m v$ .

For ball that sticks: final velocity is 0. Change in momentum: ${Δp}_{2} = m (0 - v) = m v$ . So $I_{2} = m v$ .

Comparing: $I_{2} = \frac{I_{1}}{2}$ .

Final answer: I₂=I₁/2