Question

A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question part Part 1 (a) If the ball was in contact with the floor for 0.5 ms (0.5 × 10-3 s), what was the average magnitude of the force exerted on the ball by the floor?

Answer 1

Answer:

7.5

Explanation:

Answer 2

Answer:

The  the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

Explanation:

Given:

Mass of ball $m = 0.4$ kg

Initial speed $v_{i} = -8 \frac{m}{s}$

Rebound speed $v_{f} = 6 \frac{m}{s}$

Contact time interval $\Delta t = 0.5 \times 10^{-3}$ sec

For finding the average magnitude of the force on the ball by the floor is given by,

   $F_{avg} = \frac{\Delta P}{\Delta t}$

Here $\Delta P = m (v_{f}- v_{i} )$

   $F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}$

   $F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }$

   $F_{avg} = 11.2 \times 10^{3}$ N

Therefore, the  the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N