Explanation:
It is given that,
Mass of the ball, m = 0.4 kg
Initial speed of the ball, u = 4 m/s
Final speed of the ball, v = 1 m/s
Time of contact of the ball and the floor, [tex]t=2\ ms=2\times 10^{-3}\ s[/tex]
(a) Let F is the force exerted on the ball by the floor. It can be calculated using the definition of impulse. It is given by :
[tex]J=F\times t=m(v-u)[/tex]
[tex]F=\dfrac{m(v-u)}{t}[/tex]
[tex]F=\dfrac{0.4\ kg\times (1\ m/s-4\ m/s)}{2\times 10^{-3}\ s}[/tex]
F = -600 N
|F| = 600 N
(b) The magnitude of the gravitational force that the Earth exerts on the ball is equal to the weight of the ball. It is given by :
[tex]W=mg[/tex]
[tex]W=0.4\ kg\times 9.8\ m/s^2[/tex]
W = 3.92 kg
Hence, this is the required solution.
The average magnitude of the force exerted on the ball by the floor is 600 N.
The gravitational force that Earth exerts on the ball is 3.92 N.
The given parameters;
The average magnitude of the force exerted on the ball by the floor is calculated as follows;
[tex]F = \frac{m(v-u)}{t} \\\\F = \frac{0.4(1-4)}{2.0\times 10^{-3}} \\\\F = -600 \ N\\\\|F| = 600 \ N[/tex]
The gravitational force that Earth exerts on the ball is calculated as follows;
F = mg
F = (0.4 x 9.8)
F = 3.92 N
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