Physics M.C.Q (1 Marks)

$x=2 t-1$

$v=\frac{d x}{d t}=2 \mathrm{~m} \mathrm{~s}^{-1}$

$P=F \cdot v$

$=2 \times 5=10 \mathrm{~W}$

$A$ $\rightarrow v_1$    ($B$)

                  rest

It undergoes completely inelastic collision

Using conservation of linear momentum

Initial momentum = Final momentum

$\Rightarrow m v_1=m v_2+m v_2$

$\Rightarrow m v_1=2 m v_2$

$\Rightarrow \frac{v_1}{v_2}=\frac{2}{1}$

$\text { for } x =2$

$U =\frac{1}{2} k (2)^2........(1)$

$U ^{\prime}=\frac{1}{2} k (8)^2..........(2)$

$\text { Eq. (2)/eq. (1) }$

$\Rightarrow \frac{U^{\prime}}{U^{\prime}}=\left(\frac{8}{2}\right)^2$

$\Rightarrow U^{\prime}=16\,U$

$\Rightarrow T = W + f$

$=20000+3000$

$=23000\,N$

$\text { Power } = Tv$

$=23000 \times 1.5$

$=34500 \text { watts }$

$=36 \times 10^{7}\,J$

$\mathrm{P}_{\text {input }}=\frac{\mathrm{mgh}}{\mathrm{t}}$

$=\frac{15 \times 10 \times 60}{1}=9000=9\, \mathrm{~kW}$

$10 \,\% \operatorname{loss}=0.9 \times 10^{3}$

$\mathrm{P}_{\text {output }}=9 \times 10^{3}-0.9 \times 10^{3}=8.1 \,\mathrm{~kW}$

at given point

$\mathrm{KE}=3 \mathrm{PE}$

So, $4 \mathrm{PE}=\mathrm{mgs}$

$4 \mathrm{mgh}=\mathrm{mgs}$

$\mathrm{H}=\mathrm{s} / 4$

$\mathrm{KE}=\frac{3 \mathrm{mgs}}{4}=\frac{1}{2} \mathrm{mV}^{2}$

$\mathrm{~V}=\sqrt{\frac{3\times 2 \mathrm{gs}}{4}}=\sqrt{\frac{3 \mathrm{gS}}{2}}$

$T - mg =\frac{ m (7 gr )}{ r }$

$T =8 mg$

$=0.625 \times 10^{-1}$

$=0.0625\, eV$

$\Rightarrow \mathrm{W}=\int_{0}^{1}(20+10 \mathrm{y}) \mathrm{dy}$

$\Rightarrow \mathrm{W}=20[\mathrm{y}]_{0}^{1}+10\left[\frac{\mathrm{y}^{2}}{2}\right]_{0}^{1}$

$\Rightarrow \mathrm{W}=25\; \mathrm{J}$

$\Rightarrow \bar{v}=-\frac{v}{3} \hat{i}--\frac{v}{3} \hat{j} \quad|\bar{v}|=\sqrt{2} \frac{v}{3}$

Energy released $=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(3 m)\left(\frac{2 v^{2}}{9}\right)=\frac{4}{3} m v^{2}$

At $x=8\;;\;130=\frac{1}{2}\left(\frac{1}{2}\right) v^{2}$

$\Rightarrow v=2 \sqrt{130}=22.8 \mathrm{ms}^{-1}$

For $\mathrm{v}=20.6 \mathrm{ms}^{-1}$

$\Rightarrow m(20 \hat{\imath}+25 \hat{\jmath}-12 \hat{k})=\frac{m}{6}(100 \hat{\imath}+35 \hat{\jmath}+8 \hat{k})+\frac{5 m}{6} \bar{v}_{2} $

$\Rightarrow \bar{v}_{2}=4 \hat{\imath}+23 \hat{\jmath}-16 \hat{k}$

$T$ will be maximum when $\theta=0^{\circ}$

When mass is at lowest point.

Fraction of energy lost $=\frac{\frac{1}{2}(4 m) u^{2}-\frac{1}{2}(4 m)\left(\frac{u}{3}\right)^{2}}{\frac{1}{2}(4 m) u^{2}}$

$=1-\frac{1}{9}=\frac{8}{9}$

$v_{A}=\sqrt{5 g R}$

using energy conservation $m g h=\frac{1}{2} m v_{A}^{2}$

$h=\frac{1}{2} \frac{v_{A}^{2}}{g}=\frac{1}{2} \frac{5 g}{g} \frac{D}{2} \quad\left(R=\frac{D}{2}\right)$

$h=\frac{5 D}{4}$

$\begin{gathered}
  mv + 4m \times 0 = 4mv' + 0 \Rightarrow v' = v/4 \hfill \\
  Coefficient\,of\,restitution, \hfill \\
  e = \frac{{\operatorname{Re} letive\,velocity\,of\,separation}}{{\operatorname{Re} letive\,velocity\,of\,separation}} \hfill \\
  \,\,\, = \,\frac{{v/4}}{v} = 0.25 \hfill \\ 
\end{gathered} $

Kinetic energy, $E=\frac{p^{2}}{2 m}$

$E \propto \frac{1}{m}$

$\frac{E_{1}}{E_{2}}=\frac{m_{2}}{m_{1}}=\frac{3 M}{2 M}=\frac{3}{2}$

$\frac{E_{1}}{E_{1}+E_{2}}=\frac{3}{3+2}=\frac{3}{5}$

$\frac{E_{1}}{E}=\frac{3}{5}$

$E_{1}=\frac{3 E}{5}$

$\begin{gathered}
  \left( {ii} \right)The\,total\,work\,done\,by\,gravitational\, \hfill \\
  force\,and\,the\,resistive\,force\,of\,air\,is\, \hfill \\
  equal\,to\,change\,in\,kinetic\,energy\,of\, \hfill \\
  rain\,drop. \hfill \\
  \therefore \,{W_g} + {W_r} = \frac{1}{2}m{v^2} - 0 \hfill \\
  10 + {W_r} = \frac{1}{2} \times {10^{ - 3}} \times 50 \times 50\,or\, \hfill \\
  {W_r} =  - 8.75\,J \hfill \\ 
\end{gathered} $

$\frac{L^2}{T^3}=constant$

$L \propto T^{3 / 2}$

displacement $d \propto t^{3 / 2}$

$T_{A}-m g=\frac{m v_{c}^{2}}{R}$
Energy at point $A=\frac{1}{2} m v_{0}^{2}$   $...(i)$

Energy at point $C$ is

$\frac{1}{2} m v_{c}^{2}+m g \times 2 R \ldots .$         $...(ii)$
Applying Newton's second law at point $C$

$T_{c}+m g=\frac{m v_{c}^{2}}{R}$
To complete the loop $T_{c} \geq 0$

So, $m g=\frac{m v_{c}^{2}}{R}$

$\Rightarrow \quad v_{c}=\sqrt{g R}$      $...(ii)$

From Eqs. $(i)$ and $(ii)$ by conservation of energy

$\frac{1}{2} m v_{0}^{2}=\frac{1}{2} m v_{c}^{2}+2 m g R$

$\Rightarrow \quad \frac{1}{2} m v_{0}^{2}=\frac{1}{2} m g R+2 m g R \quad\left(\because v_{c}=\sqrt{g R}\right)$

$\Rightarrow \quad v_{0}^{2}=g R+4 g R$

$\Rightarrow \quad v_{0}=\sqrt{5 g R}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,{(KE + PE)_{{\mathop{\rm Re}\nolimits} ference}} = {\left( {KE + PE} \right)_h}\\
 \Rightarrow \,\frac{1}{2}Mv_1^2 = Mgh\,\,\,or,\,\,\,{v_1} = \sqrt {2gh} \\
\,\,\,\,\,{v_1} = \sqrt {2 \times 10 \times 0.1}  = \sqrt 2 \,m{s^{ - 1}}\\
{\rm{Using}}\,momentum\,conservation\,principle\\
for\,block\,and\,bullet\,system,
\end{array}$

$\begin{array}{l}
{\left( {M \times 0 + mu} \right)_{Before\,collision}} = {\left( {M \times {v_1} + mv} \right)_{After\,collision}}\\
 \Rightarrow \,0.01 \times 400 = 2\sqrt 2  + 0.01 \times v\\
 \Rightarrow v = \frac{{4 - 2\sqrt 2 }}{{0.01}} = 117.15\,m{s^{ - 1}} \approx 120\,m{s^{ - 1}}
\end{array}$

$\begin{array}{l}
\overrightarrow v  = \int {\overrightarrow a dt = \int {\left( {2t\hat i + 3{t^2}\hat j} \right)dt = {t^2}\hat i + {t^3}\hat j\,m{s^{ - 1}}} } \\
\therefore \,power\,devloped\,by\,the\,force\,at\,time\,t,\\
P = \overrightarrow F .\overrightarrow v \, = \left( {2t\hat i + 3{t^2}\hat j} \right).\left( {{t^2}\hat i + {t^3}\hat j\,} \right)\\
W = \left( {2{t^3} + 3{t^5}} \right)W
\end{array}$

$\mathrm{W}_{man}=\mathrm{F}_{\mathrm{av}} \hat{\mathrm{i}} \cdot \mathrm{s} \hat{\mathrm{i}}=160 \times 20=3200 \mathrm{J}$

$\vec{F}=3 \hat{i}+4 \hat{j}$

Displacement vector $(\vec{x})=4 \hat{i}+3 \hat{j}$

$\vec{F} \cdot \vec{x}=(3 i+4 \hat{j}) \cdot(4 \hat{i}+3 \hat{j})$

$=12+12=24 \,J$

$m g-T=\frac{m g}{6}$

$\Rightarrow T=\frac{5}{6} m g$

$W=\frac{-5}{6} \operatorname{mgx}$

$T-m g=m a$

$T=m(g+a)$

$=\frac{4}{3} m g$

Work $(w)=T . h$

$=\frac{4}{3} m g h$

$w=\int d w=\int F d x$

$\mathrm{w}=\int \frac{\mathrm{K}}{\mathrm{v}} \mathrm{dx}$

$\mathrm{w}=\int \frac{\mathrm{K}}{\mathrm{dx}} \mathrm{dtdx}$

$\mathrm{w}=\int \mathrm{Kdt}$

$\mathrm{w}=\mathrm{Kt}$

$=\mathrm{mg} \sin \theta . \mathrm{vt} \cos (90-\theta)$

$\mathrm{W}_{f}=\mathrm{mg} \sin ^{2} \theta . \mathrm{vt}$

Height covered by projectile $=\frac{u^2 \sin ^2 \theta}{2 g}$

$W=-m g\left(\frac{u^2 \sin ^2 \theta}{2 g}\right)$

$=\frac{-m u^2 \sin ^2 \theta}{2}$

$\therefore $$\frac{{{s_2}}}{{{s_1}}} = {\left( {\frac{{{m_1}}}{{{m_2}}}} \right)^2} = {\left( {\frac{{200}}{{300}}} \right)^2}$ 

$⇒$ ${s_2} = {s_1} \times \frac{4}{9} = 36 \times \frac{4}{9} = 16\;m$

$\therefore \frac{1}{2}m{v^2}$=$\mu \;mgs$

 $⇒$ $s = \frac{{{u^2}}}{{2\mu \;g}} = \frac{{{{(10)}^2}}}{{2 \times (0.5) \times 10}} = 10\,m$  

Total height $= 20 × 0.25 = 5m $

Work done $= mgh = 70 × 9.8 × 5 = 3430 J$

i.e. ${m_A}{v_A} = {m_B}{v_B}$

==> $4 \times {v_A} = 8 \times 6$

==> ${v_A} = 12\;m/s$ 

Kinetic energy of other mass $A$ , = $\frac{1}{2}{m_A}v_A^2$ = $\frac{1}{2} \times 4 \times {(12)^2}= 288 J.$

In given problem $K.E.$ becomes $64\%$ of the original value.

$\frac{{{P_2}}}{{{P_1}}} = \sqrt {\frac{{{E_2}}}{{{E_1}}}} = \sqrt {\frac{{64E}}{{100E}}} = 0.8$==>${P_2} = 0.8\,P$

$\therefore {P_2} = 80\% $ of the original value.

i.e. decrease in momentum is $20\%$.

==> $2 \times 9.8 \times h = \frac{{490}}{2}$

==> $h = 12.5m.$

and $\frac{{{m_1}}}{{{m_2}}} = \frac{1}{3}$…(ii) 

by solving we get ${m_1} = 3kg$ and ${m_2} = 9kg$ 

Kinetic energy of smaller part = $\frac{1}{2}{m_1}v_1^2 = 216J$

$v_1^2 = \frac{{216 \times 2}}{3}$

==> ${v_1} = 12m/s$ 

So its momentum = ${m_1}{v_1} = 3 \times 12 = 36\;kg{\rm{ - }}{\rm{m}}/s$ 

As both parts possess same momentum therefore momentum of each part is $36\;kg{\rm{ - }}m/s$

$\frac{{{K_1}}}{{{K_2}}}\,\, = \,\,\frac{{\frac{1}{2}\,\,mv_1^2}}{{\frac{1}{2}\,\,mv_2^2}}\,\, = \,\,\frac{{v_1^2}}{{v_2^2}}\,\, = \,\,\frac{1}{2}$