If the basketball\'s mass is 0.5 kg, and it is in contact with the floor for 0.2

Question

If the basketball's mass is 0.5 kg, and it is in contact with the floor for 0.25 s during the bounce, what is the magnitude of the average net force exerted on the basketball during this bounce?
N

What is the average magnitude of the force exerted on the basketball by the floor during the collision? (Hint: What's different between the last question and this one?)
N

I have that the net force is as follows: 3i-4j+3i+3j= 6i-j.

The basketball pictured below is moving to the right as it bounces along the floor. The ball's momentum just before the collision is 3 i - 4 j m/s, and the momentum just after striking the ground is 3 i + 3 j m/s. What is the direction of the net force acting on the ball during the collision with the floor? Explain your answer. If the basketball's mass is 0.5 kg, and it is in contact with the floor for 0.25 s during the bounce, what is the magnitude of the average net force exerted on the basketball during this bounce? What is the average magnitude of the force exerted on the basketball by the floor during the collision? (Hint: What's different between the last question and this one?) I have that the net force is as follows: 3i-4j+3i+3j= 6i-j.

Explanation / Answer

There is no change in the "i" component of the velocity.
Only in the "j" component.
Change = final - initial
= +3j - (-4j) = +7j so the change in velocity, and the acceleration and force must be in the +j direction.

Ft = delta(mv)
F = delta(mv) /t
= 0.5 * 7 / 0.25 = 14 kg m/s^2

There is NO difference between Q2 and 3 as written.
The question almost certainly originally asked for the force on the FLOOR caused by the basketball.

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