Question

• A
rubber ball
50 gm.
falls from
height of 100cm
and rebounds to
50 cm height.
Find the impulse and
average force between ball and the ground if
impact time is
0.1
sec.​

Answer 1

Answer:

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

v

2

=u

2

+2gh

v

2

=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

v

2

=u

2

−2gh

0=u

2

−2×9.8×0.5

u

2

=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=

time

impulse

F=

0.1

0.225

F=2.25N

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

Explanation:

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