Question

A steel ball with mass 0.347 kg falls onto a hard floor and bounces. Its speed just before hitting the floor is 23.6 m/s and its speed just after bouncing is 12.7 m/s.

(a) What is the magnitude of the impulse of the ball?
(b) if the ball is in contact with the floor for 0.0139s, what is the magnitude of the average force exerted onthe ball by the floor?

Answer 1

(a)The magnitude of impulse of the ball is 12.5951kgm/s

(b)F=906.19N

Step-by-step explanation:

m=0.347kg , u=23.6m/s, v=12.7m/s

we will take the direction of final velocity as negative.

Impulse(ft)=m(v-u)

=0.347(-12.7-23.6)

Impulse=-12.5951 kgm/s

= 12.5951kgm/s

(b)t=0.0139s

F=m(v-u)/t

F=12.595/0.0139

F=906.19N