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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 contact 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

Respuesta :

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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