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, e its velocity of rebound.

Answer 1

given value

mass of steel ball is100g=0.1kg

height of steel ball h¹=1.8m

height of steel ballh²=1.25m

The potential energy of the ball before the fall:

p.e=mgh1

u1 =.08*9.8*1.8

=1.764j

p.e=k.e

1.764=k.e

k.e=1.764

The kinetic energy of the ball is,

K

E

=

1.764

J

The kinetic energy of the ball is also given by,

K

E

=

1

2

m

v

2

v

=

√

2

K

E

m

=

√

2

×

1.764

0.1

=

5.93

m

/

s

The velocity of hitting the plates is 5.93 m/s.

The kinetic energy of the rebound is,

K

E

=

P

E

=

m

g

h

2

=

0.1

×

9.8

×

1.21

=

1.186

J

The kinetic energy of rebound is 1.186 J.

e)

The kinetic energy of the ball is given by,

K

E=12mv2

v=√2

K

E

m

=

√

2

×

1.186

0.1

=

4.87

m

/

s

The velocity of rebound is 4.87 m/s

Answer 2

Answer:

a)potential energy = mgh

=0.1kg*1.8*9.8

=1.764J

b)When object falls potential energy equals kinetic energy so

answer is = 1.764J

c)K.E=1/2*m$v^{2}$

1.746J=1/2 * 0.1*$x^{2}$(assuming v as x)

x=$\sqrt{\frac{1.764}{0.1*0.5}$

x=5.93m/s

d)K.E = P.E

mgh=0.1*9.8*1.25=1.225J

e)K.E=1/2*m$v^{2}$

1.225J=1/2*0.1*$x^{2}$(assuming v as x)

x=$\sqrt \frac{1.225}{0.1*0.5}$

x=4.95m/s