Question

an elastic ball of mass 60 gram falls from a building of height 200 cm and rebounds to a height of 80cm calculate the impulse and the average force between the Between the ball and ground The ball remains in contact with ground for 0.2 second...
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Answer 1

Answer:

I THINK IT IS HELPFUL

Explanation:

Given that,

Ball of mass = 50 g

Height = 1 m

Rebound to height = 0.5 m

Time = 0.1 s

Now,

Initial velocity of the ball =um/s

Let the final velocity =vm/s

Using equation of motion

v2=u2+2gh

v2=0+2×9.8×0.1

v=1.96

v=1.4m/s

In the second time,

Final velocity=0m/s

Now, Using equation of motion

v2=u2−2gh

0=u2−2×9.8×0.5

u2=9.8

u=3.1m/s

Now, Impulse= change in linear momentum

=mv−m(−u)

=m(v+u)

=0.05×(1.4+3.1)

=0.225Ns

Now, the Force=Impulse/Time

F=timeimpulse

F=0.10.225

F=2.25N

Hence, the impulse is 0.225 Ns and force is 2.25 N