A steel ball of mass 0.5 kg is dropped from a height of 4.05m onto a horizontal steel slab. The collision is elastic, and the ball rebounds to its original height. If the ball is in contact with the slab for 0.002 sec, find the average force (in newton) on the ball during the impact?

Engineering

Physics

Straight Line Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

A steel ball of mass 0.5 kg is dropped from a height of 4.05m onto a horizontal steel slab. The collision is elastic, and the ball rebounds to its original height. If the ball is in contact with the slab for 0.002 sec, find the average force (in newton) on the ball during the impact?

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$< F >= m < a >= m (\frac{Δv}{Δt}) = \frac{m (v - (- v)}{Δt}$

$= \frac{2 mv}{Δt} = \frac{2 (m) (\sqrt{2 gh})}{Δt} = \frac{2 (0 .5) (\sqrt{2 (10) (4 .05})}{2 \times 10^{- 3}} = 4500 N$