Question

A ball is allowed to fall from a height of 10m .if there is 40% loss of energy due to impact ,then after one impact ball will go up to ?

Answer 1

Let us assume that the mass of the ball is 'M'.

A ball is allowed to fall from a height of 10m.

So, the initial height of the ball (h) = 10 m

Now,

Potential energy = mgh

→ Mg(10)

Also given that, if there is 40% loss of energy due to impact.

We have to find the final height of ball (h').

P.E.' = mgh'

→ Mg(h')

According to question,

→ P.E.' = 60% (P.E.)

→ Mg(h') = 60/100 (Mg × 10)

Mg, throughout cancel

→ h' = 60/10

→ h' = 6

Therefore,

If there is 40% loss of energy due to impact , then after one impact ball will go up to by 6 m.

Answer 2

Given:-

• $\text{A ball is allowed to fall from a height of 10 m.}$

• $\text{There is a loss of }40\% \text{ in energy due to impact.}$

To find:-

$\text{The final height of the ball in the second impact.}$

Solution:-

$\text{Let the mass of the ball be `m'.\\}\\\\h_1 \text{be the initial height of the ball before bouncing 10 m.}\\\\\text{Then, } PE_1=mgh_1\\\\\text{Let }h_2\ \text{be the final height of the ball after bouncing.}\\\\h_2=?\\\\\text{Then final potential energy becomes:}\\\\PE_2=mgh_2\\\\\text{Using the law of conservation of energy,}\\\\PE_2=60\%\ \text{of}\ PE_1 \text{(As 40\% energy is lost A/q.)}\\\\\implies ,mgh_2=\frac{60}{100}mgh_1 \\\\\implies h_2=0.6h_1$

$\implies h_2=0.6*10\\\\\boxed{\implies h_2=6\ m}$

∴$\bold{\underline{So, the\ height\ to\ which\ the\ ball\ will\ move\ after\ one\ impact\ is\ 6\ m. }}$