Question

A ball of mass 50gm is dropped from a height h=10m. It rebound losing 75% of its kinetic energy. If it remain in contact with the ground for∆t=0.01sec., the impulse of the impact force is:

Answer 1

Answer:

Given : m=0.05 kg h=10 m

Kinetic energy of ball just before collision E

i

=mgh=0.05×10×10=5 J

Thus initial momentum of the ball P

i

=

2mE

i

=

2(0.05)(5)

=0.707 N-s (downwards)

Kinetic energy of ball just after collision E

f

=

4

E

i

=

4

5

=1.25 J

Thus final momentum of the ball P

f

=

2mE

f

=

2(0.05)(1.25)

=0.353 N-s (upwards)

∴ Impulse I=ΔP=P

f

−(−P

i

)=0.353+0.707=1.06 N-s

Explanation:

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Answer 2

Answer:

0.625N-S

Explanation:

explanationin attachments