Question

A ball of mass 10 g falls from a height of 5 m. It rebounds from the ground to a height of 4 m.Find :
I) The initial potential energy of the ball
II) The kinetic energy of the ball just before striking the ground
III) The kinetic energy of the ball after striking the ground
IV) The loss in kinetic energy on striking the ground
( take g = 9.8 m/s Square )

Answer 1

Answer:

The answer will be 2.11 N

Explanation:

According to the problem the mass of the ball,m = 10 gm = 0.01 kg

The height from which the ball is dropped, h1 = 5 m

and the height the ball gained after touching the ground , h2 = 1.25 m

The collision time, Δ t = 0.1 s

Now we need to find the initial velocity of the ball, v1^2= u^2+2gh1

         As u = 0

Therefore,

v1 = √2gh1 = √2 x 20 x 5 = 10√2 m/s

Now the initial momentum p1 =  mv1 = 0.01 x (-10√2) = - 0.141 N.s

Now we need to find the final velocity of the ball, v2^2 = u^2+2gh2

             Therefore v2 =√2gh2 = √ 2 x 20 x 1.25 = 5√2 m/s

Therefore, the final momentum p2 = mv2 = 0.01 x 5√2 = 0.070 N.s

Therefore the force ,f = Δ p/Δ t = (0.070-(-0.141))/0.1 = 2.11 N

Answer 2

Answer:

I)The initial potential energy of the ball is 0.49 J.

ll)The kinetic energy of the ball just before striking the ground is 0.392 J.

III)The kinetic energy of the ball after striking the ground is 0.392 J

Iv)The loss in kinetic energy on striking the ground is 0.1 J.

Explanation:

Initial potential energy $= mgh$$= 10 \times {10}^{ - 3} \times 9.8 \times 5$=0.49 J

When it just before striking the ground all initial potential energy is converted into kinetic energy.

Hence kinetic energy just before striking the ground = 0.49 J.

When it rebounds to height 4 m,It's final potential energy =$10 \times {10}^{ - 3} \times 9.8 \times 4 = 0.392$J

Kinetic energy after striking converted to into final potential energy at 4m height.

Hence kinetic energy after striking on the ground is 0.392 J

Loss in kinetic energy$= 0.49 - 0.392 = 0.1$J