Question #286180

Stacy confirms that Marcus dropped the ball with a mass of 1.20 [kg]. The ball hits the floor at 2.40 [m/s] and bounces back at 2.00 [m/s]. The time of contact between the ball and the floor is 0.1 [s]. Assuming an elastic collision and the upward direction being the positive direction, what is the average force that the floor exerts on the ball during the impact?


1
Expert's answer
2022-01-10T09:15:22-0500
FavgΔt=mΔv,F_{avg}\Delta t=m\Delta v,Favg=mΔvΔt=m(vfvi)Δt,F_{avg}=\dfrac{m\Delta v}{\Delta t}=\dfrac{m(v_f-v_i)}{\Delta t},Favg=1.2 kg×(2 ms(2.4 ms))0.1 s=52.8 N.F_{avg}=\dfrac{1.2\ kg\times(2\ \dfrac{m}{s}-(-2.4\ \dfrac{m}{s}))}{0.1\ s}=52.8\ N.

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