Question

A 2 kg ball drops vertically onto a floor, hitting with a speed of 28 m/s. It rebounds
with an initial speed of 12 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball
is in contact with the floor for 40 millisecond, what is the magnitude of the average force on the
floor from the ball?

Answer 1

a) We can find the impulse from the formula:

$$J = p_f - p_i = mv_f - mv_i = m(v_f - v_i),$$ $$J = 2 kg \cdot (12 \dfrac{m}{s} - (-28 \dfrac{m}{s})) = 80 kg \cdot \dfrac{m}{s}.$$

b) We can find the  magnitude of the average force on the floor from the ball from the formula:

$$F \Delta t = \Delta p,$$ $$F \Delta t = m(v_f - v_i),$$ $$F = \dfrac{m(v_f - v_i)}{\Delta t} = \dfrac{J}{\Delta t} = \dfrac{80 kg \cdot \dfrac{m}{s}}{40 \cdot 10^{-3} s} = 2000N.$$

Answer:

a) $J = 80 kg \cdot \dfrac{m}{s}.$

b) $F = 2000N.$