Question

A 100g steel ball falls from a height of 1. 8 m onto a
metal plate and rebounds to a height of 1.25m. Find
a the p.e. of the ball before the fall (g= 10 m/2),
b its k e. as it hits the plate
c its velocity on hitting the plate,
d its k.e. as it leaves the plate on the rebound,

Answer 1

Answer:

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a.  PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

Explanation:

I hope this helped!

Answer 2

Answer:

ANSWER

we know that if ball is dropped from a height , h , velocity of particle just before hitting the ground will be v=

(

2gh)⇒v=

(

2g)

and if particle is projected with u , then it reaches a maximum height H then , u=

(

2gH)⇒u=

(

1.62g)

Now we know that, coefficient of restitution e=

u

1

−u

2

v

2

−v

1

=

−

(

2g)−0

0−

(

1.62g)

=

(

0.81)=0.9