Two identical steel cubes (masses 50 g, side 1 cm ) collide head-on face to face with a space of $$10 \mathrm{~cm} / \mathrm{s}$$ each. Find the maximum compression of each. Young's modulus for steel $$=Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$$.
$$\begin{aligned} &\begin{aligned} \text{Let} \quad m & =50 \mathrm{~g}=50 \times 10^{-3} \mathrm{~kg} \\ \text { Side } & =L=1 \mathrm{~cm}=0.01 \mathrm{~m} \\ \text { Speed } & =v=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s} \\ \text { Young's modulus } & =Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \end{aligned}\\ &\text { Maximum compression } \Delta L=\text { ? } \end{aligned}$$
Maximum compression $$\Delta L=$$ ?
In this case, all KE will be converted to PE By Hooke's law,
$$\frac{F}{A}=Y \frac{\Delta L}{L}$$
where $$A$$ is the surface area and $$L$$ is length of the side of the cube. If $$k$$ is spring or compression constant, then
$$\begin{aligned} \text { force } F & =k \Delta L \\ \therefore \quad k & =Y \frac{A}{L}=Y L \end{aligned}$$
$$\begin{aligned} & \text { Initial } K E=2 \times \frac{1}{2} m v^2=5 \times 10^{-4} \mathrm{~J} \\ & \text { Final } P E=2 \times \frac{1}{2} k(\Delta L)^2 \end{aligned}$$
$$\therefore \quad \Delta L=\sqrt{\frac{K E}{k}}=\sqrt{\frac{K E}{Y L}}=\sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.1}}=1.58 \times 10^{-7} \mathrm{~m} \quad[\because P E=K E]$$
A balloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.
Let $$m=$$ Mass of balloon
$$V=$$ Volume of balloon
$$\rho_{\mathrm{He}}=$$ Density of helium
$$\rho_{\text {air }}=$$ Density of air
Volume $$V$$ of balloon displaces volume $$V$$ of air.
So, $$V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g=m a=m \frac{d v}{d t}=$$ up thrust $$\quad \text{.... (i)}$$
$$\begin{aligned} &\text { Integrating with respect to } t \text {, }\\ &\Rightarrow \quad \begin{aligned} V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right) g t & =m v \quad \text{... (ii)}\\ \frac{1}{2} m v^2 & =\frac{1}{2} m \frac{V^2}{m^2}\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)^2 g^2 t^2 \\ & =\frac{1}{2 m} V^2\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)^2 g^2 t^2 \end{aligned} \end{aligned}$$
If the balloon rises to a height $$h$$, from $$s=u t+\frac{1}{2} a t^2$$,
We get $$h=\frac{1}{2} a t^2=\frac{1}{2} \frac{V\left(\rho_{\text {air }}-\rho_{\mathrm{He}}\right)}{m} g t^2\quad \text{.... (iii)}$$
From Eqs. (iii) and (ii),
$$\begin{aligned} \frac{1}{2} m v^2 & =\left[V\left(\rho_a-\rho_{\mathrm{He}}\right) g\right]\left[\frac{1}{2 m} V\left(\rho_{\mathrm{air}}-\rho_{\mathrm{He}}\right) g t^2\right] \\ & =V\left(\rho_{\mathrm{a}}-\rho_{\mathrm{He}}\right) g h \end{aligned}$$
Rearranging the terms,
$$\begin{array}{lc} \Rightarrow & \frac{1}{2} m v^2+V_{\rho_{\mathrm{He}}} g h=V_{\text {Pair }} h g \\ \Rightarrow & \mathrm{KE}_{\text {balloon }}+\mathrm{PE}_{\text {balloon }}=\text { Change in } \mathrm{PE} \text { of air. } \end{array}$$
So, as the balloon goes up, an equal volume of air comes down, increase in PE and KE of the balloon is at the cost of PE of air [which comes down].