A 1.9 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 9.0 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0158 s, what is the magnitude of the average force on the floor from the ball?

A) Impulse = 64.6 N.s

B) Force = 4088.6 N

Explanation:

We are given;

Mass of ball;m = 1.9 kg

Initial velocity of ball;u = -25 m/s (negative value because ball was drop downward)

Final velocity of ball;v = 9 m/s

From Newton's second law of motion, we know that;

Impulse = Change in momentum

Thus;

Impulse = final momentum - initial momentum = mv - mu

Thus;

Impulse = m(v - u)

Plugging in the relevant values, we have;

Impulse = 1.9(9 - (-25)

Impulse = 1.9(9 + 25)

Impulse = 1.9(34)

Impulse = 64.6 N.s

B) As said earlier,

impulse = 64.6 N.s

Now, we are looking for the magnitude of the average force.

Now, Impulse is also expressed as Force x time.

Thus,

Force x time = 64.6 N.s

We are given time as 0.0158 s

Thus, making Force the subject of the formula, we now have;

Force = 64.6/0.0158

Force = 4088.6 N

radiation

explanation:

heat is transmitted by radiation