A curved surface is shown in figure. The portion $$B C D$$ is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from $$A$$ which is at a slightly greater height than $$C$$.
With the surface $$A B$$, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(a) For which balls is total mechanical energy conserved?
(b) Which ball (s) can reach D?
(c) For balls which do not reach $$D$$, which of the balls can reach back $$A$$ ?
(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved.
Ball 3 is having negligible friction hence, there is no loss of energy.
(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.
(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in "wrong" sense when it reaches B. It cannot roll back to A, because of kinetic friction.
$$A$$ rocket accelerates straight up by ejecting gas downwards. In a small time interval $$\Delta t$$, it ejects a gas of mass $$\Delta m$$ at a relative speed $$u$$. Calculate KE of the entire system at $$t+\Delta t$$ and $$t$$ and show that the device that ejects gas does work $$=(1 / 2) \Delta m u^2$$ in this time interval (negative gravity).
Let $$M$$ be the mass of rocket at any time $$t$$ and $$v_1$$ the velocity of rocket at the same time $$t$$.
Let $$\Delta m=$$ mass of gas ejected in time interval $$\Delta t$$.
Relative speed of gas ejected $$=u$$.
Consider at time $$t+\Delta t$$
$$\begin{aligned} (\mathrm{KE})_t+\Delta t & =\mathrm{KE} \text { of rocket }+\mathrm{KE} \text { of gas } \\ & =\frac{1}{2}(M-\Delta m)(v+\Delta v)^2+\frac{1}{2} \Delta m(v-u)^2 \\ & =\frac{1}{2} M v^2+M v \Delta v-\Delta m v u+\frac{1}{2} \Delta m u^2 \\ (\mathrm{KE})_t & =\mathrm{KE} \text { of the rocket at time } t=\frac{1}{2} M v^2 \\ \Delta \mathrm{K} & =(\mathrm{KE})_t+\Delta t-(\mathrm{KE})_t \\ & =(M \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^2 \end{aligned}$$
Since, action-reaction forces are equal.
$$\begin{aligned} &\begin{aligned} \text { Hence, } \quad M \frac{d v}{d t} & =\frac{d m}{d t}|u| \\ \Rightarrow \quad M \Delta v & =\Delta m u \\ \Delta K & =\frac{1}{2} \Delta m u^2 \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Now, by work-energy theorem, }\\ &\begin{aligned} & \Delta K=\Delta W \\ \Rightarrow \quad & \Delta W=\frac{1}{2} \Delta m u^2 \end{aligned} \end{aligned}$$
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].