A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of $$50 \mathrm{~m} \mathrm{~s}^{-1}$$. Calculate
(a) the loss of PE of the drop.
(b) the gain in KE of the drop.
(c) Is the gain in KE equal to loss of PE? If not why?
Take, $$g=10 \mathrm{~ms}^{-2}$$.
$$\begin{aligned} \text { Given, mass of the rain drop } (m) & =1.00 \mathrm{~g} \\ & =1 \times 10^{-3} \mathrm{~kg} \end{aligned}$$
$$\begin{aligned} \text { Height of falling }(h) & =1 \mathrm{~km}=10^3 \mathrm{~m} \\ g & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$
Speed of the rain drop $$(v)=50 \mathrm{~m} / \mathrm{s}$$
(a) Loss of PE of the drop $$=m g h$$
$$\qquad=1 \times 10^{-3} \times 10 \times 10^3=10$$
$$\begin{aligned} \text { (b) Gain in KE of the drop } & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} \times 1 \times 10^{-3} \times(50)^2 \\ & =\frac{1}{2} \times 10^{-3} \times 2500 \\ & =1.250 \mathrm{~J} \end{aligned}$$
(c) No, gain in KE is not equal to the loss in its PE, because a part of PE is utilised in doing work against the viscous drag of air.
Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (figure). One of the bobs is released after being displaced by $$10^{\circ}$$ so that it collides elastically head-on with the other bob.
(a) Describe the motion of two bobs.
(b) Draw a graph showing variation in energy of either pendulum with time, for $$0 \leq t \leq 2 T$$. where $$T$$ is the period of each pendulum.
(a) Consider the adjacent diagram in which the bob $,$B$$ is displaced through an angle $$\theta$$ and released.
At $$t=0$$, suppose bob $$B$$ is displaced by $$\theta=10^{\circ}$$ to the right. It is given potential energy $$E_1=E$$. Energy of $$A, E_2=0$$. When $$B$$ is released, it strikes $$A$$ at $$t=T / 4$$. In the head-on elastic collision between $$B$$ and $$A$$ comes to rest and $$A$$ gets velocity of $$B$$. Therefore, $$E_1=0$$ and $$E_2=E$$. At $$t=2 T / 4, B$$ reaches its extreme right position when KE of $$A$$ is converted into $$P E=E_2=E$$. Energy of $$B, E_1=0$$.
At $$t=3 T / 4, A$$ reaches its mean position, when its PE is converted into $$\mathrm{KE}=E_2=E$$. It collides elastically with $$B$$ and transfers whole of its energy to $$B$$. Thus, $$E_2=0$$ and $$E_1=E$$. The entire process is repeated.
(b) The values of energies of $$B$$ and $$A$$ at different time intervals are tabulated here. The plot of energy with time $$0 \leq t \leq 2 T$$ is shown separately for $$B$$ and $$A$$ in the figure below.
| Time (t) | Energy of A ($$E_1$$) |
Energy of B ($$E_2$$) |
|---|---|---|
| 0 | E | 0 |
| T/4 | 0 | E |
| 2T/4 | 0 | E |
| 3T/4 | E | 0 |
| 4T/4 | E | 0 |
| 5T/4 | 0 | E |
| 6T/4 | 0 | E |
| 7T/4 | E | 0 |
| 8T/4 | E | 0 |
Suppose the average mass of raindrops is $$3.0 \times 10^{-5} \mathrm{~kg}$$ and their average terminal velocity $$9 \mathrm{~m} \mathrm{~s}^{-1}$$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
$$\begin{aligned} &\text { Given, average mass of rain drop }\\ &\begin{aligned} (m) & =3.0 \times 10^{-5} \mathrm{~kg} \\ \text { Average terminal velocity }=(V) & =9 \mathrm{~m} / \mathrm{s} \\ \text { Height }(h) & =100 \mathrm{~cm}=1 \mathrm{~m} \\ \text { Density of water }(\rho) & =10^3 \mathrm{~kg} / \mathrm{m}^3 \\ \text { Area of the surface }(A) & =1 \mathrm{~m}^2 \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { Volume of the water due to rain }(V) & =\text { Area } \times \text { height } \\ & =A \times h \\ & =1 \times 1=1 \mathrm{~m}^3 \\ \text { Mass of the water due to rain }(M) & =\text { Volume } \times \text { density } \\ & =V \times \rho \\ & =1 \times 10^3 \\ & =10^3 \mathrm{~kg} \\ \therefore \text { Energy transferred to the surface } & =\frac{1}{2} \mathrm{mv}^2 \\ & =\frac{1}{2} \times 10^3 \times(9)^2 \\ & =40.5 \times 10^3 \mathrm{~J}=4.05 \times 10^4 \mathrm{~J} \end{aligned}$$
An engine is attached to a wagon through a shock absorber of length 1.5 m . The system with a total mass of $$50,000 \mathrm{~kg}$$ is moving with a speed of $$36 \mathrm{~kmh}^{-1}$$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m . If $$90 \%$$ of energy of the wagon is lost due to friction, calculate the spring constant.
$$\begin{aligned} \text { Given, mass of the system } (m) & =50,000 \mathrm{~kg} \\ \text { Speed of the system }(v) & =36 \mathrm{~km} / \mathrm{h} \\ & =\frac{36 \times 1000}{60 \times 60}=10 \mathrm{~m} / \mathrm{s} \end{aligned}$$
$$\begin{aligned} \text { Compression of the spring }(x) & =1.0 \mathrm{~m} \\ \qquad \mathrm{KE} \text { of the system } & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} \times 50000 \times(10)^2 \\ & =25000 \times 100 \mathrm{~J}=2.5 \times 10^6 \mathrm{~J} \end{aligned}$$
Since, $$90 \%$$ of KE of the system is lost due to friction, therefore, energy transferred to shock absorber, is given by
$$\begin{aligned} \Delta E & =\frac{1}{2} k x^2=10 \% \text { of total KE of the system } \\ & =\frac{10}{100} \times 2.5 \times 10^6 \mathrm{~J} \text { or } k=\frac{2 \times 2.5 \times 10^6}{10 \times(1)^2} \\ & =5.0 \times 10^5 \mathrm{~N} / \mathrm{m} \end{aligned}$$
An adult weighting 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $$10 \%$$ of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.
$$\begin{aligned} \text { Given, weight of the adult }(w) & =m g=600 \mathrm{~N} \\ \text { Height of each step } & =h=0.25 \mathrm{~m} \\ \text { Length of each step } & =1 \mathrm{~m} \end{aligned}$$
$$\begin{aligned} & \text { Total distance travelled }=6 \mathrm{~km}=6000 \mathrm{~m} \\ & \therefore \quad \text { Total number of steps }=\frac{6000}{1}=6000 \end{aligned}$$
$$\begin{aligned} \text { Total energy utilised in jogging } & =n \times m g h \\ & =6000 \times 600 \times 0.25 \mathrm{~J}=9 \times 10^5 \mathrm{~J} \end{aligned}$$
Since, $$10 \%$$ of intake energy is utilised in jogging.
$$\therefore$$ Total intake energy $$=10 \times 9 \times 10^5 \mathrm{~J}=9 \times 10^6 \mathrm{~J}$$.