A ball of mass 0.1kg dropped from a building of 8m high unto a hard floor bounces back to a height of 2m. Calculate the change in momentum , if the ball is in constant with the floor of 0.1s, what is the force exerted on the body . (g = 10ms)

Answer 1

a. V1^2 = Vo^2 + 2g*h.
V1^2 = 0 + 20*8, V1 = 12.65 m/s.

V^2 = V2^2 + 2g*h.
0 = V2^2 - 20*2, V2 = 6.32 m/s.

M*V1-M*V2 = 0.1*12.65 - 0.1*6.32 = 0.633 = Change in momentum.

b. V = V2 + a*t.
0 = 6.32 - a*0.1, a = 63.2 m/s^2.

F = M*a = 0.1 * 63.2 = 6.32 N.

Answer 2

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