Physics: A1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s. (a) what impulse acts on the ball during the contact? (b) if the ball is in contact with the floor for 0.020 s, what is the magnitude of the average force on the floor from by the ball? neglect air resistance

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A1.2 kg ball drops vertically onto a floor from

iamPaola

11.01.2022

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 32}$

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

J=m(v-u)

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

$J=\dfrac{F}{t}$

$F=J\times t$

$F=42.048\times 0.02$

F = 0.8409 N

Hence, this is the required solution.

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