A ball whose mass is 0.1 kg hits the floor with a speed of 6 m/s and rebounds upward with a speed of 4 m/s. If the ball was in contact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor? Favg = N. Calculate the magnitude of the gravitational force that the Earth exerts on the ball: mg = N In a collision, for a brief time there are forces between the colliding objects that are much greater than external forces. Which of the following is true during the collision between the bouncing ball and the floor? The magnitude of the gravitational force exerted by the Earth is much greater than the magnitude of the force exerted by the floor. The magnitude of the force exerted by the floor is much greater than the magnitude of the gravitational force exerted by the Earth. The force exerted by the floor and the gravitational force exerted by the Earth are approximately equal in magnitude. Additional Materials

Let's start by analyzing the change in linear momentum in the collision of the ball with the floor. The magnitude of the initial momentum carried by the ball is the product of its mass times its initial velocity:

[tex]p_i=m\,v_i\\p_i=0.1\,(6 )\,\frac{kg\,m}{s} \\p_i=0.6\,\frac{kg\,m}{s}[/tex]

The magnitude of the ball's final momentum immediately after the collision with the floor is:

[tex]p_f=m\,v_i\\p_f=0.1\,(4 )\,\frac{kg\,m}{s} \\p_f=0.4\,\frac{kg\,m}{s}[/tex]

a) To calculate the magnitude of the change in momentum, we need to have in mind that the initial momentum was pointing towards the center of the earth, while the final one was pointing in the opposite direction, therefore their difference as vectors must include the change in direction, which ends up giving an addition of their magnitudes:

Change in momentum: [tex]= (0.6 + 0.4 )\frac{kg\,m}{s} = 1\,\frac{kg\,m}{s}[/tex]

Then the magnitude of the force exerted by the floor on the ball is this change in momentum divided by the time of the impact:

[tex]F=\frac{\Delta\,p}{\Delta\,t} = \frac{1}{1.5\,10^{-3}} N=667\,N[/tex]

b) The magnitude of the Earth's gravitational force on the object is:

[tex]F_g=m\,g=0.1\,(9.8)\,N=0.98\,N[/tex]

Therefore, "The magnitude of the force exerted by the floor is much greater than the magnitude of the gravitational force exerted by the Earth."

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-30T10:50:38+02:00

The magnitude of the force exerted by the floor is much greater than the magnitude of the gravitational force exerted by the Earth.

Calculating the forces exerted on the ball:

The mass of the ball is m = 0.1 kg.

The speed of the ball just before the rebound from the floor is u = 6m/s

The speed of the ball just after the rebound from the floor is v = -4m/s

time for which the ball is in contact with the floor is t = 1.5ms = 0.0015s

Force is defined as the rate of change in the momentum of the object given by:

F = Δp/t

Δp is the change in momentum, and t is the time taken

Δp = m(u-v) = 0.1(6-(-4)) = 1 kgm/s

So, F = 1/0.0015

F = 667N

The gravitational force exerted on the ball is given by:

F' = mg = 0.1×9.8

F' = 0.98N

Comparing F and F' we get:

The magnitude of the force exerted by the floor is much greater than the magnitude of the gravitational force exerted by the Earth.

Learn more about gravitational force:

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