Motion

For Case I :

$\mathrm{a}_{\mathrm{rel}}=0$

For collision $\frac{\mathrm{v}_0}{\sqrt{2}}=\frac{\mathrm{v}}{\sqrt{2}}$

$\begin{aligned} & \therefore \mathrm{v}=\mathrm{v}_0 \\ & \text { So } \mathrm{T}_1=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{\sqrt{2}}+\frac{\mathrm{v}}{\sqrt{2}}} \\ & \therefore \tau_1=\frac{\mathrm{L}}{\sqrt{2} \mathrm{v}_0} \quad \text{... (1)} \end{aligned}$

For case II,

$\mathrm{a}_{\mathrm{rel}}=0$

For collision, $\frac{v_0 \sqrt{3}}{2}=\frac{v}{2}$

$\begin{aligned} & \therefore \mathrm{v}=\sqrt{3} \mathrm{v}_0 \\ & \text { So, } \mathrm{T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\mathrm{v} \frac{\sqrt{3}}{2}} \\ & \mathrm{~T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\frac{3 \mathrm{v}_0}{2}} \\ & \therefore \mathrm{T}_2=\frac{\mathrm{L}}{2 \mathrm{v}_0} \quad \ldots . .(2) \\ & \text { so, }\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=(\sqrt{2})^2=2 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=2 \end{aligned}$

Given that $\frac{d x_1}{d t}=$ speed of person $=60 \mathrm{~cm} / \mathrm{s}$



Also $\frac{d x_2}{d t}=$ speed of tip of person's shadow

Applying similar triangle rule in $\triangle A B E $ and $ \triangle D C E$

$ \begin{aligned} & \frac{4}{x_2}=\frac{1.6}{x_2-x_1} \\\\ & 4 x_2-4 x_1=1.6 x_2 \\\\ & 2.4 x_2=4 x_1 \end{aligned} $

Differentiate both sides w.r.t. $t$

$ \begin{aligned} & 2.4 \frac{d x_2}{d t}=4 \frac{d x_1}{d t} \\\\ & \frac{d x_2}{d t}=\frac{4}{2.4}(60) \\\\ &=100 \mathrm{~cm} / \mathrm{s} \end{aligned} $

This is the speed of the tip of the person's shadow with respect to the lamp post. But, we need to find the speed of the shadow's tip with respect to the person, which is the relative speed :

$ \begin{aligned} \vec{v}_{S P} & =\vec{v}_{S G}-\vec{v}_{P G} \\\\ v_{S P} & =100 \mathrm{~cm} \mathrm{~s}^{-1}-60 \mathrm{~cm} \mathrm{~s}^{-1} \\\\ & =40 \mathrm{~cm} \mathrm{~s}^{-1} \end{aligned} $

$d=\frac{u^{2} \sin 2 \theta}{g}$

$ H=\frac{u^{2} \sin ^{2} \theta}{2 g} $

So, after entering in the new region, time taken by projectile to reach ground

$ \begin{aligned} t &=\sqrt{\frac{2 H}{g^{\prime}}} \\\\ &=\sqrt{\frac{2 u^{2} \sin ^{2} \theta \times 0.81}{2 g \times g}} \\\\ &=\frac{0.94 \sin \theta}{g} \end{aligned} $

So, horizontal displacement done by the projectile in new region is

$ \begin{aligned} & x=\frac{0.9 u \sin \theta}{g} \times u \cos \theta \end{aligned} $

$ \begin{aligned} x &=\frac{0.9 u \sin \theta}{g} \times u \\\\ &=0.9 \frac{u^{2} \sin 2 \theta}{2 g} \end{aligned} $

So, $d^{\prime}=\frac{d}{2}+x$

$ =0.95 d $

So, $n=0.95 d$

For first projectile,

$ < V > = {R \over T} = {U_x} = {v_1}$

For journey,

$ < V{ > _{1\,to\,n}} = {{{R_1} + {R_2} + ... + {R_n}} \over {{T_1} + {T_2} + ... + {T_n}}}$

$ = {{{{2{u_{{x_1}}} + 2{u_{{y_1}}}} \over g} + {{2{u_{x2}} + 2{u_{{y_2}}}} \over g} + ... + {{2{u_{{x_n}}} + 2{u_{{y_n}}}} \over g}} \over {{{2{u_{{y_1}}}} \over g} + {{2{u_{y2}}} \over g} + ...{{2{u_{{y_n}}}} \over g}}}$

${u_x}\left[ {{{1 + {1 \over {{\alpha ^2}}} + {1 \over {{\alpha ^4}}} + ...{1 \over {{\alpha ^{2n}}}}} \over {1 + {1 \over \alpha } + {1 \over {{\alpha ^2}}} + ...{1 \over {{\alpha ^n}}}}}} \right] = 0.8{v_1}$

${{{v_0}\left[ {{1 \over {1 - {1 \over {{\alpha ^2}}}}}} \right]} \over {\left[ {{1 \over {1 - {1 \over \alpha }}}} \right]}} = 0.8{v_1}$

$ \Rightarrow {\alpha \over {1 + \alpha }} = 0.8$

$ \Rightarrow \alpha = 4$

$H = {{{u^2}{{\sin }^2}(45^\circ )} \over {2g}} = 120$ m

$ \Rightarrow {{{u^2}} \over {4g}} = 120$ m

If speed is v after the first collision, then speed should remain ${1 \over {\sqrt 2 }}$ times, because kinetic energy has reduced to half.

$ \Rightarrow v = {u \over {\sqrt 2 }}$

$ \Rightarrow {h_{\max }} = {{{v^2}{{\sin }^2}(30^\circ )} \over {2g}}$

$ \Rightarrow {h_{\max }} = {{{{\left( {{u \over {\sqrt 2 }}} \right)}^2}{{\sin }^2}(30^\circ )} \over {2g}}$

$ \Rightarrow {h_{\max }} = \left( {{{{u^2}/4g} \over 4}} \right) = {{120} \over 4}$

$ \Rightarrow {h_{\max }} = 30$ m

Consider the motion of two balls with respect to rocket.

Distance travelled by ball A from left end of the chamber is

$ = {{{u^2}} \over {2a}} = {{{{(0.3)}^2}} \over {2 \times 2}} = {{0.09} \over 4} \approx 0.02$ m

So, collision of two balls will take place very near to left end of the chamber.

For ball B

$S = ut{1 \over 2}a{t^2}$

$ - 4 = - 0.2 \times t - {1 \over 2} \times 2 \times {t^2}$

${t^2} + 0.2t - 4 = 0$

$t = {{ - 0.2 \pm \sqrt {{{(0.2)}^2} - 4(1)( - 4)} } \over 2} = {{ - 0.2 \pm \sqrt {0.04 + 16} } \over 2}$

t = 1.9 s, $-$2.1 s

Since t can't be negative

$\therefore$ t = 1.9 s

Nearest integer is 2 s.

Also, from

$S = ut + {1 \over 2}a{t^2}$

${S_A} = 0.3t + {1 \over 2}( - 2){t^2} = 0.3t - {t^2}$

${S_B} = 0.2t + {1 \over 2}(2){t^2} = 0.2t + {t^2}$

$\because$ ${S_A} + {S_B} = 4$

$\Rightarrow$ 0.5 t = 4 or t = 8 s

Since A observes B as moving normal to it

${v_B}\cos 30^\circ = {v_A}$

${v_B}{{\sqrt 3 } \over 2} = 100\sqrt 3 $

${v_B} = 200$ m/s

Therefore, ${t_0} = {{500} \over {200\sin 30^\circ }} = 5$ s

u = 10 ms^$-$1, $\theta$ = 60$^\circ$

Time of flight is

$t = {{2u\sin \theta } \over g} = {{2 \times 10 \times \sin 60^\circ } \over {10}} = \sqrt 3 \,s$

Let v be the velocity of the train. The horizontal velocity of ball at the instant it is thrown $ = (v + {u_x}) = (v + u\cos \theta )$. Therefore, the horizontal range of the ball with respect to the ground is

$R = (v + u\cos \theta )t$, where $t = \sqrt 3 \,s$

It is clear that

Distance travelled by ball in time $t + 1.15 = R$

i.e. $vt + {1 \over 2}a{t^2} + 1.15 = (v + u\cos \theta )t$

$ \Rightarrow {1 \over 2}a{t^2} + 1.15 = (u\cos \theta )t$

$ \Rightarrow {1 \over 2}a \times {(\sqrt 3 )^2} + 1.15 = (10\cos 60^\circ ) \times \sqrt 3 $

$ \Rightarrow a = 5$ ms^$-$2

From the diagram, you can see a body is present at two different position A and B at the same time which is not possible.