Question

A 240 g ball is dropped from a height of 2.1 m, bounces on a hard floor, and rebounds to a height of 1.6 m. The figure shows the impulse received from the floor. What maximum force does the floor exert on the ball?
impulse= .005*Fmax

Answer 1

FΔt=mΔv−−>F=m(Δv)/(Δt)
We can find the speed by calculating the potential energy and converting it to kinetic energy. Ep=Ek−−>mgh=mv^2/2

v=(√2gh)=(√2∗9.81∗2.1)≈6.42m/s

so then F=.240∗6.42/0.005=308 N (we know the time because of the above mentioned impulse equation.)

Answer 2

The maximum force exerted by the floor is 380N.

Impulse:

Impulse is the vector change in instantaneous momentum. Mathematically, the magnitude of impulse is defined as:

I =  FΔt = mΔv

where F is the force applied and Δt is the time of impact or change

According to the question:

I = 0.005×F

⇒Δt =0.005

Now the potential energy of the ball when it is at a height of 2.1 m is:

PE = mgh = 0.240×9.8×2.1

PE = 4.94 J

when the ball reaches the ground the potential energy is converted into kinetic energy:

KE = (1/2)mv² = 4.94 J

0.240v² = 2×4.94

v = 6.42 m/s , or

Δv = 6.42 m/s

Now we know that:

FΔt = mΔv

F = mΔv/Δt

F = (0.240×6.42)/0.005

F ≈ 380 N

Learn more about impulse:

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