Centre Of Mass And Collision Practice Questions

$m_1 = 3kg,m_2 = 1kg$Mass of plate-1 is assumed to be concentrated at (0.5, 1.5)Mass of plate-2 is assumed to be concentrated at (1.5, 2.5).$x_\text{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$ $={3\times 0.5+1\times 1.5}/4=0.75$ $y_\text{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$ $={3\times 1.5+1\times 2.5}/4=1.75$

Angular momentum conservation ${mv} l= \frac{ {M} l^{2}}{3} \omega + {m} l^{2} \omega$ $\Rightarrow \omega = \frac{1 \\times 6 \\times 1}{\ \frac{2}{3}+1}= \frac{18}{5}$ Now using energy consevation $\frac{1}{2}({M} \frac{l^{2}}{3} ) \omega ^{2}+ \frac{1}{2}({m} l^{2} ) \omega ^{2}$ $=( {m}+ {M}) {r}_{ {cm}}(1-\cos \theta )$ $= ({m}+ {M}) ({{m} l+\frac{{M} l}{2}}/{ {m}+ {M}} )$ ${g}(1-\cos \theta )$ $\frac{5}{6} \\times (\frac{18}{5} )^{2}=20(1-\cos \theta )$ $\Rightarrow 1-\cos \theta = \frac{18}{5}\times \frac{3}{20}$ $\cos \theta =1- \frac{27}{50}$ $\cos \theta = \frac{23}{50} \Rightarrow \theta \ ≃ 63°$

Applying law of conservation of linear momentum &0.1 \times 20=(1.9+0.1) V $2=2 V$ $V=1 m / sec$ $KE =(1 / 2) mv ^{2}$ $=(1 / 2) 2(1)^{2}$ $=1 J$ $TE = KE + mgh$ $=1+2 \times 10 \times 1$ $=21 J$

Particles will collide after time $t_0=h/{\sqrt{2gh}}$ at collision, $v_A = gt_0$, $v_B = u_B - gt_0$ $\Rightarrow v_A=-v_B$ Time taken by combined mass to reach the ground $\text{time} =\sqrt{{2\times 3h/4}/g}=\sqrt{\frac{3h}{2g}}$

By conservation of linear momentum :$(0.1)(3\hat{i})+(0.1)(5\hat{j})$ $=(0.1)(4)(\hat{i}+\hat{j})+(0.1)\vec{v}$ $\Rightarrow \vec{v}=-\hat{i}+\hat{j}$∴ Speed of B after collision $|\vec{v}|=\sqrt2$Now, kinetic energy $=1/2mv^{2-}1/2(0.1)(2)=1/10$ $\therefore x=1$

By momentum conservation,${\mu}/2+\mu=2mv'$ $v'={3v}/4$Range after collision $={3v}/4\sqrt{ {2H}/g}$ $={3v}/4\sqrt{ \frac{2 u^2\sin^2 60°}{g . 2g} }$ $=\frac{3\sqrt3u^2}{8g}$

Let 1 kg as origin and x-y axis as shown $x_{cm}={1(0)+1.5(3)+2.5(0)}/5$ $=0.9\text{cm}$ $y_{cm}={1(0)+1.5(0)+2.5(4)}/5$ $=2\text{cm}$

${{\vec{v}}}_{01}=(\sqrt{3} \hat{i}+\hat{j}) {\text{m}} / {\text{s}}$ ${\vec{v}}_{02}=\vec{0}$ ${m}_{1}=2 {m}_{2}$ After collision, v_\vec{1}=(\hat{i}+\sqrt3\hat{j})\text{m}/\text{s} v_\vec{2}=\? Applying conservation of linear momentum, ${m}_{1} {\vec{v}}_{01}+{m}_{2} {\vec{v}}_{02}={m}_{1} {{\vec{v}}}_{1}+{m}_{2} {{\vec{v}}}_{2}$ $2 {m}_{2}(\sqrt{3} \hat{i}+\hat{j})+0=2 {m}_{2}(\hat{i}+\sqrt{3} \hat{j})+{m}_{2} {{\vec{v}}}_{2}$ ${{v}}_{2}=2(\sqrt{3} \hat{i}+\hat{j})-2(\hat{i}+\sqrt{3} \hat{j})$ $=2(\sqrt{3} {\hat{i}}-{\hat{j}})+2({\hat{i}}-\sqrt{3} {\hat{j}})$ ${{\vec{v}}}_{2}=2(\sqrt{3}-1)(\hat{i}-\hat{j})$ for angle between ${{\vec{v}}}_{1} \& {{\vec{v}}}_{2}$, $\cos \theta ={{{\vec{v}}}_{1}.{{\vec{v}}}_{2}}/{{{\vec{v}}}_{1} {{\vec{v}}}_{2}}$ $=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 \times 2 \sqrt{2}(\sqrt{3}-1)}$ $\cos \theta =\frac{1-\sqrt{3}}{2 \sqrt{2}}\Rightarrow \theta =105°$or

$x=\frac{3 R}{8}=3 {cm}$ $x=3$

From momentum conservation ${\vec{P}}_{ {i}}={\vec{P}}_{{f}}$ ${m}( {ui})+3{m}(0)= {mvj}+3 {m} \overline{{v}}_{1}$ ${mui}- {mvj}=3 \ {m} \overline{{v}}_{1}$ $\overline{{v}}_{1}=\frac{{ui}- {v} {j}}{3}$ or $|{v}_{1} |={\sqrt{ {u}^{2}+ {v}^{2}}}/{3}$ or ${v}_{1}^{2}=\frac{{u}^{2}+ {v}^{2}}{9}$ .....(1) As collision is perfectely elastic hence ${k}_{ {i}}= {k}_{ {j}}$ $\frac{1}{2} {\mu}^{2}+ \frac{1}{2} 3 {m} 0^{2}$ $= \frac{1}{2} {mv}^{2}+ \frac{1}{2} 3 {mv}_{1}^{2}$ $\Rightarrow {u}^{2}= {v}^{2}+3 {v}_{1}^{2}$ ${u}^{2}= {v}^{2}+3 \frac{({u}^{2}+ {v}^{2} )}{9}$ $\Rightarrow 3 {u}^{2}=3 {v}^{2}+ {u}^{2}+ {v}^{2}$ $\Rightarrow2 {u}^{2}=4 {v}^{2}$ $\ {v}=\frac{{u}}{\\sqrt{2}}$

By concept of COM$m_1R_1 = m_2R_2$Remaining mass × (2 - R) = cavity mass × (R - 1)$(4/3\pi R^3\rho -4/3\pi 1^3\rho )(2-R)$ $=4/3\pi 1^3\rho \times (R-1)$ $(R^3 - 1) (2 - R) = R - 1$ $(R^2 + R + 1) (2 - R) = 1$

${X}_{ {\text{com}}}={{m}_{1} {x}_{1}- {m}_{2} {x}_{2}}{ {m}_{1}- {m}_{2}}$ where : ◆ $\ {m}_{1}=$ mass of complete disc ◆ ${m}_{2}=$ removed mass ◆ Let $\sigma =$ surface mass density of disc material wrt 'O' : ${X}_{ {\text{com} }}= {\sigma \pi {a}^{2}( {O})-\sigma .\frac{{a}^{2}}{4} .{d}} /{\sigma \pi {a}^{2}-\sigma \frac{{a}^{2}}{4}}$ $= {-\frac{{a}^{2}}{4} {d}}/{\pi {a}^{2}-\frac{{a}^{2}}{4}}$ $=\frac{- {d}}{4 \pi -1}=-\frac{{a}}{2(4 \pi -1)}$ So, $X=2(4 \pi -1)$ $=(8 \pi -2)=23.12$ So, nearest integer value of $X=23$

Momentum conservation along x $2 {mv}_{0} \cos \theta =2 {m} \frac{{v}_{0}}{2}$ $\cos\theta =1/2$ $\theta =60°$ Angle is 2θ = 120°

$x_\text{cm}=\frac{∫xdm}{∫dm}=\frac{∫(\lambda dx)x}{∫dm}$ $=\frac{∫^{L}_{0}(a+{bx^2}/L^2)xdx}{∫^{L}_{0}(a+{bx^2}/L^2) dx}$ $=\frac{ {aL^2}/2+b/L^2 .L^4/4 }{aL+b/L^2 . L^3/3}$ $=\frac{( {4a+2b}/8)L}{(3a+b)/3}=3/4\frac{(2a+b)L}{(3a+b)}$

By momentum conservation along $y:$ $\ {m}_{1} {u}_{1} \sin \theta _{1}= {m}_{2} {u}_{2} \sin \theta _{2}$ i.e. ${\mu}_{1} \sin \theta _{1}=10 {\mu}_{2} \sin \theta _{2}$ \Rightarrow $[u_1\sin\theta _1=10u_2\sin\theta _2]$ ......(i) ${kf}_{\ {m}_{1}}= \frac{1}{2} {ki}_{ {m}_{1}}$ i.e. $\frac{1}{2} {\mu}_{1}^{2}={1}{2}\times \frac{1}{2} {\mu}^{2}$ i.e. $[u_{1}= \frac{u}{\\sqrt{2}}]$ ........(ii) Also collision is elastic $: {k}_{ {i}}= {k}_{ {f}}$ $\frac{1}{2} {\mu}^{2}= \frac{1}{2} {\mu}_{1}^{2}+ \frac{1}{2} .10 {m} .{u}_{2}^{2}$ $\frac{1}{2} {\mu}^{2}= \frac{1}{2} \times \frac{1}{2} {\mu}^{2}$ $+ \frac{1}{2} \\times 10 {m}.{u}_{2}^{2}$ $\frac{1}{4} {\mu}^{2}= \frac{1}{2} \\times 10 \\times {\mu}_{2}^{2}$ $[{u}_{2}=\frac{{u}}{\\sqrt{20}}]$ ......(iii) Putting(ii) & (iii) in (i) $\frac{ {u}}{\\sqrt{2}} \sin \theta _{1}=10.\frac{{u}}{\\sqrt{20}} \sin \theta _{2}$ $[\sin \theta _{1}=\sqrt{10} \sin \theta _{2} ] \to$ Hence $\ {n}=10$

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the final velocity be ${v}',$ so on applying momentum conservation: ${mv}=16 {m} {v}'\Rightarrow {v}'= {v} / 16$ Now initial energy ${E}_{ {i}}= \frac{1}{2} \ {mv}^{2}$ Final energy $: {E}_{ {f}}= \frac{1}{2} \\times 16 \ {m} \times (\frac{{v}}{16} )^{2}$ $\Rightarrow {E}_{ {f}}= \frac{1}{2} {m} \frac{{v}^{2}}{16}$ Energy loss : ${E}_{ {i}}- {E}_{ {f}}$ $\Rightarrow \frac{1}{2} {mv}^{2}- \frac{1}{2} {m} \frac{{v}^{2}}{16}$ $\Rightarrow \frac{1}{2} {mv}^{2} [1- \frac{1}{16} ] \Rightarrow \frac{1}{2} {mv}^{2}[{15} /{16} ]$ $\% {p}= \frac{\text{Energy loss}}{\text{ Original energy }} \\times 100$ $=\frac{ \frac{1}{2} {mv}^{2}[\frac{15}{16} ]}{ \frac{1}{2} {mv}^{2}} \\times 100$ $=93.75 \%$ $\Rightarrow$ Value of $\ {P}$ is close to 94.

$\therefore r_{1}=(\;\frac{\;\frac{n}{2}}{m+\;\frac{m}{2}}) l$ $\;r_{1}=\;\frac{l}{3}$ When rod is rotated by $\theta _{0}$ and released, then it look this. After roating $\theta _{0}$ and released, when the rod go through mean position it will have maximum speed. ${V}_{ {max}}= {A} \omega$ $\;\; = {r}_{1} \theta _{0} \omega$ $\;\; =(\;\frac{l}{3} \theta _{0}) \omega$ We know, $\omega =\sqrt{\;\frac{K}{I}}$ $\;\text{ and }\; {I}=m(\;\frac{l}{3})^{2}+\;\frac{m}{2}(\;\frac{2 l}{3})^{2}$ $\;\; =\;\frac{m l^{2}}{9}+\;\frac{4 m l^{2}}{18}$ $\;\; =\;\frac{2 m l^{2}+4 m l^{2}}{18}$ $\;\; =\;\frac{m l^{2}}{3}$ $\therefore \omega =\sqrt{\;\frac{3 K}{m l^{2}}}$ $\therefore$ Tension in the rod when it passes through the mean position is, ${T}=\;{m {v^{2}}_{\max }}/{r_{1}}$ $=\;\frac{m(\frac{l}{3} \theta _{0} \omega )^{2}}{\frac{l}{3}}$ $=\;\frac{m(\frac{l}{3})^{2} \theta _{0}^{2} \omega ^{2}}{\frac{l}{3}}$ $=m(\;\frac{l}{3}) \theta _{0}^{2}(\;\frac{3 K}{m l^{2}})$ $=\;\frac{K \theta _{0}^{2}}{l}$

$P_{i}=P_{f}$ $4 m v=2\ ( \frac{m v_{1}}{\\sqrt{2}}\ )$ $v_{1}=2 \\sqrt{2} v$

$mv_0$ = $mv_2-mv_1$ $1/2 mV_1^2$ = 0.36 × $1/2 mV_0^2$ $v_1$ = 0.6 $v_0$ $1/2 MV_2^2$ = 0.64 × $1/2 mV_0^2$ $V_2$ = $\sqrt{m/M}$ × 0.8 $V_0$ $mV_0$ = $\sqrt{mM}$ × $0.8 V_0$ - m × $0.6V_0$ \Rightarrow 1.6 m = 0.8 $\sqrt{mM}$ $4m^2$ = mM