A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81 \over 100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms - 2)

Motion in A Straight Line PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q55 with Solution

View written solutionFree

Correct answer: A

Given data

Initial height: $h_0 = 5\,\text{m}$
Rebound ratio: $r = \dfrac{81}{100} = 0.81$
Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

We need the average speed over the complete motion, i.e.

\text{Average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}

Total distance travelled

The ball first falls through $5$ m.

Then it rises to $5r$ , falls again through $5r$ , rises to $5r^2$ , falls through $5r^2$ , and so on.

So,

D = 5 + 2\left(5r + 5r^2 + 5r^3 + \cdots \right)

D = 5 + 10\left(r + r^2 + r^3 + \cdots \right)

Using the geometric series,

r + r^2 + r^3 + \cdots = \frac{r}{1-r}

Hence,

D = 5 + 10\cdot \frac{r}{1-r}

Substitute $r=0.81$ :

D = 5 + 10\cdot \frac{0.81}{0.19}

D = 5 + 10\cdot \frac{81}{19} = 5 + \frac{810}{19} = \frac{95+810}{19} = \frac{905}{19}

D \approx 47.63\,\text{m}

Total time taken

For a fall or rise through height $h$ , the time is

t = \sqrt{\frac{2h}{g}}

First fall

t_0 = \sqrt{\frac{2\cdot 5}{10}} = \sqrt{1} = 1\,\text{s}

Subsequent up and down motions

If the ball rises to height $5r^n$ , then time to go up is

\sqrt{\frac{2(5r^n)}{10}} = \sqrt{r^n} = r^{n/2}

and time to come down is the same.

So total time is

T = 1 + 2\left(\sqrt{r} + r + r^{3/2} + \cdots \right)

This is a geometric series with first term $\sqrt{r}$ and common ratio $\sqrt{r}$ .

Thus,

\sqrt{r} + r + r^{3/2} + \cdots = \frac{\sqrt{r}}{1-\sqrt{r}}

Now,

\sqrt{r} = \sqrt{0.81} = 0.9

Therefore,

T = 1 + 2\cdot \frac{0.9}{1-0.9}

T = 1 + 2\cdot \frac{0.9}{0.1} = 1 + 18 = 19\,\text{s}

Average speed

\text{Average speed} = \frac{D}{T} = \frac{905/19}{19} = \frac{905}{361}

\text{Average speed} \approx 2.51\,\text{m s}^{-1}

So the nearest option is

\boxed{2.50\,\text{m s}^{-1}}

Option check

A: $2.50\,\text{m s}^{-1}$ ✅
B: $3.0\,\text{m s}^{-1}$ ❌
C: $2.0\,\text{m s}^{-1}$ ❌
D: $3.50\,\text{m s}^{-1}$ ❌

Hence the correct option is A.

More from Motion in A Straight Line

View written solutionFree

Correct answer: A

Given data

Initial height: $h_0 = 5\,\text{m}$
Rebound ratio: $r = \dfrac{81}{100} = 0.81$
Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

We need the average speed over the complete motion, i.e.

\text{Average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}

Total distance travelled

The ball first falls through $5$ m.

Then it rises to $5r$ , falls again through $5r$ , rises to $5r^2$ , falls through $5r^2$ , and so on.

So,

D = 5 + 2\left(5r + 5r^2 + 5r^3 + \cdots \right)

D = 5 + 10\left(r + r^2 + r^3 + \cdots \right)

Using the geometric series,

r + r^2 + r^3 + \cdots = \frac{r}{1-r}

Hence,

D = 5 + 10\cdot \frac{r}{1-r}

Substitute $r=0.81$ :

D = 5 + 10\cdot \frac{0.81}{0.19}

D = 5 + 10\cdot \frac{81}{19} = 5 + \frac{810}{19} = \frac{95+810}{19} = \frac{905}{19}

D \approx 47.63\,\text{m}

Total time taken

For a fall or rise through height $h$ , the time is

t = \sqrt{\frac{2h}{g}}

First fall

t_0 = \sqrt{\frac{2\cdot 5}{10}} = \sqrt{1} = 1\,\text{s}

Subsequent up and down motions

If the ball rises to height $5r^n$ , then time to go up is

\sqrt{\frac{2(5r^n)}{10}} = \sqrt{r^n} = r^{n/2}

and time to come down is the same.

So total time is

T = 1 + 2\left(\sqrt{r} + r + r^{3/2} + \cdots \right)

This is a geometric series with first term $\sqrt{r}$ and common ratio $\sqrt{r}$ .

Thus,

\sqrt{r} + r + r^{3/2} + \cdots = \frac{\sqrt{r}}{1-\sqrt{r}}

Now,

\sqrt{r} = \sqrt{0.81} = 0.9

Therefore,

T = 1 + 2\cdot \frac{0.9}{1-0.9}

T = 1 + 2\cdot \frac{0.9}{0.1} = 1 + 18 = 19\,\text{s}

Average speed

\text{Average speed} = \frac{D}{T} = \frac{905/19}{19} = \frac{905}{361}

\text{Average speed} \approx 2.51\,\text{m s}^{-1}

So the nearest option is

\boxed{2.50\,\text{m s}^{-1}}

Option check

A: $2.50\,\text{m s}^{-1}$ ✅
B: $3.0\,\text{m s}^{-1}$ ❌
C: $2.0\,\text{m s}^{-1}$ ❌
D: $3.50\,\text{m s}^{-1}$ ❌

Hence the correct option is A.

Motion in A Straight Line question

Motion in A Straight Line question

First fall

Subsequent up and down motions

More from Motion in A Straight Line

Motion in A Straight Line question

Motion in A Straight Line question

First fall

Subsequent up and down motions

More from Motion in A Straight Line