Question

A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 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?

Answer 1

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

$F=\frac {m(v-u)}{t}$

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

$F=\frac {0.3(2--5)}{0.0015}=1400N$