Explanation:
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, $\frac{v_0 \sqrt{3}}{2}=\frac{v}{2}$
$\begin{aligned} & \therefore \mathrm{v}=\sqrt{3} \mathrm{v}_0 \\ & \text { So, } \mathrm{T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\mathrm{v} \frac{\sqrt{3}}{2}} \\ & \mathrm{~T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\frac{3 \mathrm{v}_0}{2}} \\ & \therefore \mathrm{T}_2=\frac{\mathrm{L}}{2 \mathrm{v}_0} \quad \ldots . .(2) \\ & \text { so, }\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=(\sqrt{2})^2=2 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=2 \end{aligned}$
Explanation:
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} $
A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$, then the new range is $d^{\prime}=n d$. The value of $n$ is ___________ .
Explanation:
$ 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, $n=0.95 d$
Explanation:
$ < 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$
Explanation:
$H = {{{u^2}{{\sin }^2}(45^\circ )} \over {2g}} = 120$ 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
Explanation:
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
Explanation:
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
Explanation:
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
STATEMENT-2: If the observer and the object are moving at velocities ${\overrightarrow v _1}$ and ${\overrightarrow v _2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is ${\overrightarrow v _2}$ - ${\overrightarrow v _1}$.
While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is :
(i) the time interval between the firings, and
(ii) the coordinates of the point P.
Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane.Is the time variation of position, shown in the figure observed in nature?
Explanation:
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.
where x is in meters and t in seconds. Find
(i) The displacement of the particle when its velocity is zero, and
(ii) The work done by the force in the first 6 seconds.
(i) maximum velocity reached, and
(ii) the total distance travelled.