Question

A rubber ball falling from a height of 5m rebounds from a hard floor to a height 3.5 cm. the %loss of energy during the impact is

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

A rubber ball falling from a height of 5 m rebounds from a hard floor to a height 3.5 m

$\checkmark$ To Find:

The %loss of energy during the impact

$\checkmark$ Solution:

We know that,

Potential Energy (P.E)

$\red{\star \ \boxed{\sf P.E=mgh}}$

Given that,

A rubber ball falling from a height of 5 m rebounds from a hard floor to a height 3.5 m

Therefore,

$\green{\implies \sf h_{1}=5 \ m}$

$\blue{\implies \sf h_{2}=3.5 \ m}$

Hence,

Potential Energy due to height 5 m

$\implies \sf (P.E)_{1}=mgh_{1}$

Potential Energy due to height 3.5 m

$\implies \sf (P.E)_{2}=mgh_{2}$

We know that,

% Energy Loss =

$\red{\sf \% \ Energy \ Loss = \left(\dfrac{(P.E)_{1}-(P.E)_{2}}{(P.E)_{1}} \right)\times 100}$

$\sf \% \ Energy \ Loss = \left(\dfrac{mgh_{1}-mgh_{2}}{mgh_{2}}\right) \times 100$

Cancelling common terms [mg] in both numerator and denominator

We get,

$\pink{\sf \% \ Energy \ Loss= \left(\dfrac{h_{1}-h_{2}}{h_{1}}\right) \times 100}$

Given that,

$\green{\implies \sf h_{1}=5 \ m}$

$\blue{\implies \sf h_{2}=3.5 \ m}$

Hence,

Substituting the values,

We get,

$\green{\sf \% \ Energy \ Loss= \left(\dfrac{5-3.5}{5}\right) \times 100}$

On further simplification,

We get,

$\sf \% \ Energy \ Loss= 0.3\times 100$

$\blue{\sf \% \ Energy \ Loss= 30}$

Hence,

$\checkmark$ Energy Loss = 30 %