Question

S SECTION
05a) A ball of mass 20g falls from a height of 20m and after striking the ground, it rebounds
to a ht. of 8m. Calculate :
9) K.E. of the ball just before striking the ground
2) As the ball rebounds, how much K.E. changes to P.E.? Take g=10 m/s2.
121​

Answer 1

Answer:

SOLVED

Explanation:

K.E. of the ball just before striking the ground

= MV²/2 = MGH = 20X10X20 = 4000 J

As the ball rebounds, how much K.E. changes to P.E.?

MV2/2 = MGH = 20X10X8 = 1600 J

K.E CHANGED TO P.E = 4000 - 1600 = 2400 J

Answer 2

The Kinetic energy of the ball just before striking the ground and kinetic changes to potential energy after rebounding the ball are 4 J and 2.4 respectively.

A ball of mass 20g falls from a height of 20m and after striking the ground, it rebounds to a height of 8m.

We have to calculate

1. Kinetic energy of the ball just before striking the ground.
2. As the ball rebounds, how much kinetic energy changes to Potential energy ?

1.  According to conservation of energy,

Potential energy of the ball at a height of 20m = kinetic energy of ball just before striking the ground.

⇒ kinetic energy of the ball just before striking the ground = mgh₁

= 20g × 10m/s² × 20m

= 0.02kg × 10 m/s² × 20m

= 4 kgm²/s²

= 4J

Therefore the kinetic energy of the ball just before striking the ground is 4 J.

2. Change in potential energy = Change in kinetic energy [ from Energy conservation theorem ]

⇒ mgh₁ - mgh₂ = change in kinetic energy

⇒ Change in kinetic energy = mg(h₁ - h₂)

= 20g × 10 m/s² × (20m - 8m)

= 0.02 kg × 10m/s² × 12 m

= 2.4 J

Therefore 2.4 J of kinetic energy changes to potential energy after rebounding the ball.

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