The bob $$A$$ of a pendulum released from horizontal to the vertical hits another bob $$B$$ of the same mass at rest on a table as shown in figure.
If the length of the pendulum is 1 m , calculate
(a) the height to which bob A will rise after collision.
(b) the speed with which bob B starts moving.
Neglect the size of the bobs and assume the collision to be elastic.
When ball $$A$$ reaches bottom point its velocity is horizontal, hence, we can apply conservation of linear momentum in the horizontal direction.
(a) Two balls have same mass and the collision between them is elastic, therefore, ball $$A$$ transfers its entire linear momentum to ball $$B$$. Hence, ball $$A$$ will come to at rest after collision and does not rise at all.
$$\begin{aligned} &\text { (b) Speed with which bob } B \text { starts moving }\\ &\begin{aligned} & =\text { Speed with which bob } A \text { hits bob } B \\ & =\sqrt{2 g h} \\ & =\sqrt{2 \times 9.8 \times 1} \\ & =\sqrt{19.6} \\ & =4.42 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$
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}$$