Question

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.6 m. (a) what fraction of its initial energy is lost during the bounce? (b) what is the ball's speed just before and just after the bounce? (c) where did the energy go?

Answer 1

Answer:

c to the ball

Explanation:

Answer 2

Answer : Part a) Fraction of energy lost : 20 %

Part b) Speed before and after bounce = 6.3 m/s and 5.6 m/s

Part c) Energy is lost as thermal energy .

Part A) Fraction of energy lost during bouncing :

The energy possessed by any object when present at any height is potential energy . The formula of potential energy is given as :

PE = mgh

where PE = potential energy

,m = mass pf object , g = gravitational acceleration and h = height

Given : Initial height , h₁ = 2 m final height , h₂ = 1.6 m

Initial potential energy : m * g* h ₁

Final potential energy = m* g* h₂

Energy lost = Initial PE - Final PE

= ( mgh₁ - mgh2 )

Fraction of energy lost : $\frac{energy lost}{initial energy}=\frac{mgh1 - mgh2}{mgh1}$

Plugging value in above formula and taking " mg " common =>

Fraction of energy lost = $\frac{mg(h1-h2)}{mg (h1)} * 100$

$= \frac{(h1-h2)}{h1} * 100$

$= \frac{(2 - 1.6) }{2} * 100$

Fraction of energy lost = 20%

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Part B ) Speed of ball just before and after the bounce.

Speed of ball before the bounce :

The potential energy gets converted to kinetic energy when it fall from height of 2m , so

Potential energy = kinetic energy

mgh₁ = $\frac{1}{2}$ m v²

or v ² = 2gh₁

Given : g = 9.8 m/s² h= 2 m

v² = 2 * 9.8 m/s² * 2 m = 39.2 m²/s²

v = 6.3 m/s

Speed of ball after bounce :

Potential energy = kinetic energy

mgh₂ = $\frac{1}{2}$ m v²

or v² = 2gh₂

= 2 * 9.8 m/s² * 1.6 m = 31.36 m²/s²

v = 5.6 m/s

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Part C) The energy lost due to friction. When the ball touches the ground , there occur friction force between the surface of ground and ball , due to which energy is lost as thermal energy .