(a) Let $$v_1$$ and $$v_2$$ are velocities of the two balls after collision. Now, by the principle of conservation of linear momentum,

$$\begin{aligned} 2 m v_0 & =m v_1+m v_2 \\ \text{or}\quad 2 v_0 & =v_1+v_2 \\ \text{and}\quad e & =\frac{v_2-v_1}{2 v_0} \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & v_2=v_1+2 v_0 e \\ \therefore & 2 v_1=2 v_0-2 e v_0 \\ \therefore & v_1=v_0(1-e) \end{array}$$

Since, $$e<1 \Rightarrow v_1$$ has the same sign as $$v_0$$, therefore, the ball moves on after collision.

(b) Consider the diagram below for a general collision.

By principle of conservation of linear momentum,

$$P=P_1+P_2$$

For inelastic collision some KE is lost, hence $$\frac{p^2}{2 m}>\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}$$

$$\therefore \quad p^2>p_1^2+p_2^2$$

Thus, $$p, p_1$$ and $$p_2$$ are related as shown in the figure.

$$\theta$$ is acute (less than 90$$^\circ$$) ($$p^2=p_1^2+p_2^2$$ would given $$\theta=90^{\circ}$$)

$$\begin{aligned} &\text { We know that }\\ &\begin{aligned} & & \text { Total energy } E & =P E+K E \\ \Rightarrow & & E & =V+K\quad \text{... (i)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { For region A Given, } V>E \text {, From Eq. (i) }\\ &K=E-V \end{aligned}$$

$$\begin{aligned} &\text { as }\\ &V>E \Rightarrow E-V<0 \end{aligned}$$

Hence, $$K<0$$, this is not possible.

For region B Given, $$V0$$

This is possible because total energy can be greater than PE $$(V)$$.

For region C Given, $$K>E \Rightarrow K-E>0$$

from Eq. (i) $$\mathrm{PE}=V=E-K<0$$

Which is possible, because PE can be negative.

For region D Given, $$V>K$$

This is possible because for a system PE (V) may be greater than KE (K).

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.

$$\begin{aligned} &\text { (b) Speed with which bob } B \text { starts moving }\\ &\begin{aligned} & =\text { Speed with which bob } A \text { hits bob } B \\ & =\sqrt{2 g h} \\ & =\sqrt{2 \times 9.8 \times 1} \\ & =\sqrt{19.6} \\ & =4.42 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { Given, mass of the rain drop } (m) & =1.00 \mathrm{~g} \\ & =1 \times 10^{-3} \mathrm{~kg} \end{aligned}$$

$$\begin{aligned} \text { Height of falling }(h) & =1 \mathrm{~km}=10^3 \mathrm{~m} \\ g & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$

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.

(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.