A 200 g ball dropped from height od 50 cm rebounds ro a height of 35 cm after im

Question

A 200 g ball dropped from height od 50 cm rebounds ro a height of 35 cm after impact.If the interaction with the floor lasted 0.01 second, what was the force felt by the ball?

Explanation / Answer

Impulse equation Ft = mv-mu To rebound to 0.35m, ball has to overcome gravity, so this allows you to calculate the speed that it left the ground at: v² = u²+2as v=0 as it finishes here, and note vectors =>a becomes -10m/s² So, u²=2as u = v(2as) = v(2x10x0.35)= ~2.64m/s For ball to be at rest on ground, and accelerated back up again, Ft = mv-mu mv is the momentum of the ball as it LEAVES the ground - the initial case of "u" for correct notation: mv = 0.2kg (NOTE kg and units!) x 2.64 mv = 0.48 kgm/s NOW!! Ball also hits ground with velocity: v²=u²+2as v²=0²+2x10x0.5 v²=10 v=v10~3.16m/s This then becomes u as initial velocity DOWN Ft = mv-mu Using +ve direction as UP, mv = + 0.632 kgm/s and mu = - 0.528 kgm/s So, Ft=0.632 -(-) 0.528 = 1.16 kgm/s Now, t = 1x10^(-2) seconds (again, units) SO F= 116N

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