Question

A stone is dropped from a height $h$. It hits the ground with a certain momentum $p$. If the same stone is dropped from a height $100\%$ more than the previous height, the momentum when it hits the ground is changed by ________?

Answer 1

Step 1: Given Data

Given: Momentum $=p$;

Let the mass of the ball be$‘m’$.

The height is incremented by $100\%$ with respect to the previous height:

If the previous height was ‘$h$’, the new height $H$ becomes ‘$2h$’.

Step 2: Case 1: When the ball is dropped from a height ‘$h$’

Initial velocity ($u_1$) $=0$

Let the final velocity be $v_1$

Net momentum $=p=m{v}_1$

Step 3: Case 2: When the ball is dropped from a height $H$,

Initial velocity ($u_2$)$=0$

Let the final velocity be ${\mathrm{v}}_2$

Net momentum $=p^{\text{'}}=m{\mathrm{v}}_2\left(\mathrm{i}\right)$

Step 4: Establish the Equation of Proportionality

From the third equation of uniformly accelerated motion:

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

$$\begin{gathered} {{\mathrm{v}}_2}^2={{\mathrm{u}}_2}^2+2\mathrm{gH} \\ {{\mathrm{v}}_2}^2={\left(0\right)}^2+2\mathrm{gH} \\ {{\mathrm{v}}_2}^2=2\mathrm{gH} \\ {\mathrm{v}}_2=\sqrt{2\mathrm{gH}}(\mathrm{equation}(\mathrm{ii}\left)\right) \end{gathered}$$

Replacing the ${\mathrm{v}}_2$by $\sqrt{2\mathrm{gH}}$in equation (i)

We get: $p^{\text{'}}=\mathrm{m}\sqrt{2\mathrm{gH}}$

We can derive from the equation that $p^{\text{'}}\propto \sqrt{\mathrm{H}}$

In general, we can say that: $p\propto \sqrt{h}$

Step 5: Finding the relationship between momentum in case 1 and case 2:

$$\begin{gathered} \frac{p^{\text{'}}}{p}=\sqrt{\frac{\mathrm{H}}{h}} \\ \frac{p^{\text{'}}}{p}=\sqrt{\frac{2h}{h}} \\ \frac{p^{\text{'}}}{p}=\sqrt{2}=1.414 \\ {p}^{\text{'}}=1.414p \end{gathered}$$

Step 6: Calculating the percentage change in the momentum:

Percentage change$=\frac{(p^{\text{'}}-p)}{p}\times 100=\frac{(1.414p-p)}{p}\times 100=41.4\%$

Therefore, the momentum when it hits the ground is changed by $41.4\%$.