Balancing frictional force for centripetal force $$\frac{m v^2}{r}=f=\mu N=\mu m g$$ where, $$N$$ is normal reaction.

$$\therefore \quad v=\sqrt{\mu r g}\quad$$ (where, $$r$$ is radius of the circular track)

For path ABC $$\quad$$ Path length

$$\begin{aligned} & =\frac{3}{4}(2 \pi 2 R)=3 \pi R=3 \pi \times 100 \\ & =300 \pi \mathrm{m} \end{aligned}$$

$$\begin{aligned} v_1=\sqrt{\mu 2 R g} & =\sqrt{0.1 \times 2 \times 100 \times 10} \\ & =14.14 \mathrm{~m} / \mathrm{s} \\ \therefore \quad t_1 & =\frac{300 \pi}{14.14}=66.6 \mathrm{~s} \end{aligned}$$

$$\begin{aligned} \text { For path } D E F \quad \text { Path length } & =\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi \\ v_2 & =\sqrt{\mu R g}=\sqrt{0.1 \times 100 \times 10}=10 \mathrm{~m} / \mathrm{s} \\ t_2 & =\frac{50 \pi}{10}=5 \pi \mathrm{s}=15.7 \mathrm{~s} \end{aligned}$$

For paths, $$C D$$ and $$F A$$

$$\begin{aligned} \text { Path length } & =R+R=2 R=200 \mathrm{~m} \\ t_3 & =\frac{200}{50}=4.0 \mathrm{~s} \end{aligned}$$

$$\therefore$$ Total time for completing one round

$$t=t_1+t_2+t_3=66.6+15.7+4.0=86.3 \mathrm{~s}$$

(a) Displacement vector of the particle of mass $$m$$ is given by

$$\mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t$$

$$\therefore \quad$$ Displacement along $$x$$-axis is,

$$\begin{aligned} x & =A \cos \omega t \\ \text{or}\quad\frac{x}{A} & =\cos \omega t \quad \text{... (i)} \end{aligned}$$

Displacement along $$y$$-axis is,

and $$\quad y=B \sin \omega t$$

or $$\quad \frac{y}{B}=\sin \omega t$$

Squaring and then adding Eqs. (i) and (ii), we get

$$\frac{x^2}{A^2}+\frac{y^2}{B^2}=\cos ^2 \omega t+\sin ^2 \omega t=1$$

This is an equation of ellipse.

Therefore, trajectory of the particle is an ellipse.

$$\begin{aligned} &\text { (b) Velocity of the particle }\\ &\begin{aligned} \mathbf{v} & =\frac{d \mathbf{r}}{d t}=\hat{\mathbf{i}} \frac{d}{d t}(A \cos \omega t)+\hat{\mathbf{j}} \frac{d}{d t}(B \sin \omega t) \\ & =\hat{\mathbf{i}}[A(-\sin \omega t) \cdot \omega]+\hat{\mathbf{j}}[B(\cos \omega t) \cdot \omega] \\ & =-\hat{\mathbf{i}} A \omega \sin \omega t+\hat{\mathbf{j}} B \omega \cos \omega t \end{aligned} \end{aligned}$$

$$\text { Acceleration of the particle }(\mathbf{a})=\frac{d \mathbf{v}}{d t}$$

$$\begin{aligned} \text{or}\quad\mathbf{a} & =-\hat{\mathbf{i}} A \omega \frac{d}{d}(\sin \omega t)+\hat{\mathbf{j}} B \omega \frac{d}{d t}(\cos \omega t) \\ & =-\hat{\mathbf{i}} A \omega[\cos \omega t] \cdot \omega+\hat{\mathbf{j}} B \omega[-\sin \omega t] \cdot \omega \\ & =-\hat{\mathbf{i}} A \omega^2 \cos \omega t-\hat{\mathbf{j}} B \omega^2 \sin \omega t \\ & =-\omega^2[\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t] \\ & =-\omega^2 \mathbf{r} \end{aligned}$$

$$\begin{aligned} &\therefore \text { Force acting on the particle, }\\ &F=m \mathbf{a}=-\mathbf{m} \omega^2 \mathbf{r}, \end{aligned}$$

Hence proved.

(a) When ball is given only horizontal velocity Horizontal velocity at the time of release $$\left(u_x\right)=v_s$$

During projectile motion, horizontal velocity remains unchanged,

$$\begin{aligned} \text{Therefore, } \quad & v_x=u_x=v_s \\ \text{In vertical direction,} \quad & v_y^2=u_y^2+2 g H \\ v_y=\sqrt{2 g H} \quad \left(\because u_y=0\right) \end{aligned}$$

$$\begin{aligned} &\therefore \text { Resultant speed of the ball at bottom, }\\ &\begin{aligned} v & =\sqrt{v_x^2+v_y^2} \\ & =\sqrt{v_s^2+2 g H} \quad \text{... (i)} \end{aligned} \end{aligned}$$

$$\text { (b) When ball is given horizontal velocity and a small downward velocity }$$

Let the ball be given a small downward velocity $$u$$.

$$\begin{aligned} \text{In horizontal direction} \quad & v_x^{\prime}=u_x=v_s \\ \text{In vertical direction } \quad & v_y^{\prime 2}=u^2+2 g H \\ \text{or} \quad & v_y^{\prime}=\sqrt{u^2+2 g H} \end{aligned}$$

$$\therefore$$ Resultant speed of the ball at the bottom

$$v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^2+u^2+2 g H} \quad \text{... (ii)}$$

From Eqs. (i) and (ii), we get $$\quad v^{\prime}>v$$

Consider the adjacent diagram, in which forces are resolved.

On resolving forces into rectangular components, in equilibrium forces $$\left(F_1+\frac{1}{\sqrt{2}}\right) N$$ are equal to $$\sqrt{2} N$$ and $$\boldsymbol{F}_2$$ is equal to $$\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right) \mathrm{N}$$.

$$\begin{array}{rlrl} \therefore & F_1+\frac{1}{\sqrt{2}} & =\sqrt{2} \\ & & F_1 & =\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{2-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.707 \mathrm{~N} \\ & \text { and } & F_2 & =\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}} \mathrm{~N}=\frac{3}{\sqrt{2}} \mathrm{~N}=2.121 \mathrm{~N} \end{array}$$

(a) Consider the adjacent diagram, force of friction on the box will act up the plane.

$$\begin{aligned} &\text { For the box to just starts sliding down } m g\\ &\begin{array}{ll} & \sin \theta=f=\mu N=\mu m g \cos \theta \\ \text { or } \quad & \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu) \end{array} \end{aligned}$$

(b) When angle of inclination is increased to $$\alpha>\theta$$, then net force acting on the box, down the plane is

$$\begin{aligned} F_1 & =m g \sin \alpha-f=m g \sin \alpha-\mu N \\ & =m g(\sin \alpha-\mu \cos \alpha) . \end{aligned}$$

(c) To keep the box either stationary or just move it up with uniform speed, upward force needed, $$F_2=m g \sin \alpha+f=m g(\sin \alpha+\mu \cos \alpha)$$ (In this case, friction would act down the plane).

(d) If the box is to be moved with an upward acceleration $$a$$, then upward force needed, $$F_3=m g(\sin \alpha+\mu \cos \alpha)+m a$$.