Question

A body dropped freely from a height $\mathrm{h}$ onto a horizontal plane, bounces up and down, and finally comes to rest. The coefficient of restitution is $\mathrm{e}$, the ratio of distance traveled in two consecutive rebounds

• A. $1:\mathrm{e}$
• B. $\mathrm{e}:1$
• C. ${\mathrm{e}}^2:1$
• D. $1:{\mathrm{e}}^2$

Answer 1

The correct option is D

$1:{\mathrm{e}}^2$

Step 1. Given data

The body is dropped from a $\mathrm{h}$, so its initial velocity is $0$ and gravity will act on it in the downward direction.

We have to find the distance travelled in two consecutive rebounds.

Step 2. Formula to be used.

According to the equation of motion,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$

Here, $\mathrm{v}$ is the velocity at any point in time, $\mathrm{u}$ is the initial velocity, $\mathrm{a}$ is the acceleration and $\mathrm{s}$ is the distance travelled.

So,

${\mathrm{v}}^2-{\left(0\right)}^2=2\times \mathrm{g}\times \mathrm{h}$

$\mathrm{v}=\sqrt{2\mathrm{gh}}$

Step 3. Calculate the ratio of distance travelled in two consecutive rebounds

According to the given, after the body hits the ground, its velocity decreases as shown in the above figure.

So,

$\mathrm{v}\text{'}=\mathrm{e}\times \mathrm{v}$

The body will have a velocity $\mathrm{ev}$ after the first collision.

Now, after rebounding for the second time, its velocity is,

${\mathrm{v}}^{\mathrm{n}}=\mathrm{e}\times \left(\mathrm{ev}\right)$

${\mathrm{v}}^{\mathrm{n}}={\mathrm{e}}^2\mathrm{v}$

Now, the ratio of distance in beginning to distance travel after two rebounds is:

$\frac{\mathrm{d}}{{\mathrm{d}}^{\mathrm{n}}}=\frac{\sqrt{2\mathrm{gh}}}{{\mathrm{e}}^2\sqrt{2\mathrm{gh}}}$

$\frac{\mathrm{d}}{{\mathrm{d}}^{\mathrm{n}}}=\frac{1}{{\mathrm{e}}^2}$

Hence, option $\mathrm{D}$ is the correct answer.