A ball whose mass is 0.4 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c ontact with the floor for 2 ms (2multiply10-3 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
1 answer:
Answer:
Explanation:
We shall apply concept of impulse
Impulse = Force x time
= Force x 2 x 10⁻³ N.s
impulse = change in momentum
change in momentum
= .4 x 4 - ( - .4 x 2 )
= 2.4
Force x 2 x 10⁻³ = 2.4
Force = 2.4 / 2 x 10⁻³
= 1.2 x 10³ N .
average magnitude of the force exerted by floor = 1.2 x 10³ N
If R be reaction force by earth
R - mg = 1.2 x 10³
R = 1.2 x 10³ + mg
= 1200 + .4 x 9.8
= 1200 +3.92
= 1203 .92 N .
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