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 ________?

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Solution

## Step 1: Given DataGiven: 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 ProportionalityFrom the third equation of uniformly accelerated motion: ${v}^{2}={u}^{2}+2\mathrm{as}$${{\mathrm{v}}_{2}}^{2}={{\mathrm{u}}_{2}}^{2}+2\mathrm{gH}\phantom{\rule{0ex}{0ex}}{{\mathrm{v}}_{2}}^{2}={\left(0\right)}^{2}+2\mathrm{gH}\phantom{\rule{0ex}{0ex}}{{\mathrm{v}}_{2}}^{2}=2\mathrm{gH}\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{2}=\sqrt{2\mathrm{gH}}\left(\mathrm{equation}\left(\mathrm{ii}\right)\right)$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:$\frac{{p}^{\text{'}}}{p}=\sqrt{\frac{\mathrm{H}}{h}}\phantom{\rule{0ex}{0ex}}\frac{{p}^{\text{'}}}{p}=\sqrt{\frac{2h}{h}}\phantom{\rule{0ex}{0ex}}\frac{{p}^{\text{'}}}{p}=\sqrt{2}=1.414\phantom{\rule{0ex}{0ex}}{p}^{\text{'}}=1.414p$Step 6: Calculating the percentage change in the momentum:Percentage change$=\frac{\left({p}^{\text{'}}-p\right)}{p}×100=\frac{\left(1.414p-p\right)}{p}×100=41.4%$Therefore, the momentum when it hits the ground is changed by $41.4%$.

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