Centre Of Mass And Collision Practice Questions

The situation is shown below Using conservation of linear momentum in $y$-direction, $\;\text{ As, }\;\;p_{i}\;=p_{f}$ $\;\text{ and }\;\;p_{i}\;=0$ $\Rightarrow \;p_{f}\;=m v_{1} \sin 30^{\circ}-m v_{2} \sin 30^{\circ}$ $\;0\;=m \times \;\frac{1}{2} v_{1}-m \times \;\frac{1}{2} v_{2}$ $\Rightarrow \;v_{1}\;=v_{2} \;\text{ or }\; v_{1}: v_{2}=1: 1$ $\;\text{ Since, }\; v_{1}: v_{2}=\;x: y\;\;\text{ (gi ven) }\;$ $\therefore \;x=1$

COM of semi-circular ring is at $\frac{2 {R}}{\pi }$ Distance from centre ⇒ x = 2

Impulse = change in momentum Ball (a) $|{\Delta{p}}^{\to }|=2 {\mu}={J}_{1}$ Ball (b) $|{{\Deltap}}^{\to }|=2 {\mu} \cos 45°={J}_{2}$ $\frac{{J}_{1}}{{J}_{2}}=\frac{1}{\cos 45°}=\\sqrt{2}$

Given, radius of hole, $r= {a} / 2$ and radius of disc, $R= {a}$ Let $x_{C M}$ be the centre of mass of system, $m_{1}, x_{1}$ be the mass and centre of mass of disc $m_{2}, x_{2}$ be the mass and centre of mass of circular hole. $\therefore \;m_{2}=\;\frac{m_{1}}{\pi R^{2}} \pi r^{2}$ $\Rightarrow \;m_{2}=\;\frac{m_{1} r^{2}}{R^{2}}=\;\frac{m_{1}(a / 2)^{2}}{a^{2}}=\;\frac{m_{1}}{4}$ and $x_{ {CM}}=\;\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$ $\therefore \;\; {x}_{ {CM}}=\;\frac{m_{1} {a}-(m_{1} / 4)(3 {a} / 2)}{m_{1}-(m_{1} / 4)}$ $\Rightarrow \;\; {x}_{ {CM}}=\;\frac{m_{1} {a}(1-3 / 8)}{3 m_{1} / 4} \Rightarrow {x}_{ {CM}}=\;\frac{5 {a}}{6}$

Given, mass of body 1 = M Mass of body 2 = m Initial speed of body $1, u_{1}=v_{0}$ Initial speed of body $2, u_{1}=0$ Final speed of body 1 and $2=v_{1}$ and $v_{2}$ Angle made by body 1 and 2 after collision with respect to initial direction $=\theta _{1}, \theta _{2}$ By using law of conservation of momentum, Along X-axis $\text{Mv}_{0}+m .0=\text{M} v_{1} \cos \theta _{1}+\text{mv}_{2} \cos \theta _{2}$ If $\theta _{1}=\theta _{2}=\theta$ $\therefore \text{Mv}_{0}=\text{Mv}_{1} \cos \theta +m v_{2} \cos \theta$ ...(i) Along Y-axis, $\text{Mv}_{1} \sin \theta _{1}=m v_{2} \sin \theta _{2}$ [since, $\theta _{1}=\theta _{2}=\theta$] $v_{2}=\frac{\text{Mv}_{1}}{m}$...(ii) Substituting the value of $v_{2}$ in Eq. (i), we get $\text{Mv}_{0}=\text{Mv}_{1} \cos \theta +m(\frac{\text{Mv}_{1}}{m}) \cos \theta$ $=2 \text{Mv}_{1} \cos \theta$ $v_{1}=\frac{v_{0}}{2 \cos \theta }$ ...(iii) By using law of conservation of energy along X-axis $\frac{1}{2} \text{Mv}_{0}^{2}=\frac{1}{2} \text{Mv}_{1}^{2}+\frac{1}{2} mv_{2}^{2}$ $\Rightarrow \text{Mv}_{0}^{2}=\text{Mv}_{1}^{2}+m(\frac{\text{M}}{m} v_{1})^{2}$ [from Eq. (ii)] $\Rightarrow \text{Mv}_{0}^{2}=\text{Mv}_{1}^{2}+\frac{\text{M}^{2} v_{1}^{2}}{m}={\text{Mv}_{1}^{2}}{m}(m+\text{M})$ $\Rightarrow v_{0}^{2}=(\frac{v_{0}}{2 \cos \theta })^{2}(\frac{m+\text{M}}{m})$ [From Eq. (iii)] $\Rightarrow v_{0}^{2}=\frac{v_{0}^{2}}{4 \cos ^{2} \theta }(1+\frac{\text{M}}{m})$ $\Rightarrow 4 \cos ^{2} \theta =1+\frac{\text{M}}{m}$ $\Rightarrow \frac{\text{M}}{m}=4 \cos ^{2} \theta -1$ [ For largest possible value of $\frac{\text{M}}{m}, \theta =0$ ] $=4 \cos ^{2} 0°-1=4-1=3$

The given situation is shown belowInitial velocity of the projectile, $u=u \cos 45^{\circ} {i}^{∧}+u \sin 45^{\circ} {j}^{∧}$ Final velocity of the projectile, $v=v \cos 45^{\circ} {i}^{∧}-v \sin 45^{\circ} {j}^{∧}$ At point $B, v \cos 45^{\circ}=u \cos 45^{\circ}$ and $v \sin 45^{\circ}=u \sin 45^{\circ}$ $v=u \cos 45^{\circ} {i}^{∧}-u \sin 45^{\circ} {j}^{∧}$ The magnitude of change in the momentum between the points $A$ and $B$ $|\Delta p|=|m(v-u)|$ $\Rightarrow \;\; \Delta p=2 m u \sin 45^{\circ}$ $\Rightarrow \;\; \Delta p=2 \times 0.005 \times 5 \sqrt{2} \times \;\frac{1}{\sqrt{2}}$ $\Rightarrow \;\; \Delta p=5 \times 10^{-2} {kg}- \frac{m}{s}$ Hence, the value of the $x$ is 5 .

By momentum conservation $4 \\times 10^{-3}(50-{v})-4 {v}=0$ ${v}=\frac{4 \\times 10^{-3} \\times 50}{4+4 \\times 10^{-3}} \ \approx 0.05 {ms}^{-1}$ Impulse ${J}={mv}=4 \\times .05=0.2 {kgms}^{-1}$

Given Mass of large block of wood, $M=5.99 {kg}$ Mass of bullet, $m=10 {g}$ Height at which their centre of mass rise, $h=9.8 {cm}$ From the law of conservation of energy, Energy of the system when bullet gets embedded = Energy of the system till it momentarily comes to rest. $\Rightarrow \;\; \;\frac{1}{2}(M+m) v_{1}^{2}=(M+m) g h$ where, $v_{1}=$ velocity of bullet $+$ block system $\Rightarrow \;\; v_{1}=\sqrt{2 g h}$.....(i) According to law of conservation of momentum, Momentum before collision $=$ Momentum after collision. $\Rightarrow \;\; m v=(M+m) v_{1}$ [where, $v=$ velocity of bullet before collision] $\Rightarrow \;\; m v=(M+m) \sqrt{2 g h}$ [using Eq. (i)] $\Rightarrow \;\; v=(\;\frac{M+m}{m}) \sqrt{2 g h}$ $\Rightarrow \;\; v=\;\frac{(5.99+0.01)}{10 \times 10^{-3}} \times \sqrt{2 \times 9.8 \times 9.8 \times 10^{-2}}$ $\Rightarrow \;\; v=831.55 {ms}^{-1}$

We can represent the given situation in figure asIt means linear momentum is conserved along $X$-axis. According to the law of conservation of linear momentum, $p_{i}=p_{f}$ m_{A} u_{A}+m_{B} u_{B}=m_{A} v_{A}+(m_{B} v_{B} \cos θ) $=m_{A} v_{A}+m_{B_{1}} v_{B_{1}} \cos 90^{\circ}+m_{B_{2}} v_{B_{2}} \cos \theta$ $\Rightarrow 10 \times 10 \sqrt{3}+20 \times 0=10 \times 0+10 \times 10 \times 0+10 \times 20 \cos \theta$ $\Rightarrow \;\; 10 \times 10 \sqrt{3}=200 \cos \theta$ where, $\cos \theta$ being the horizontal component i.e., along $X$-axis $\Rightarrow \;\; \cos \theta =\sqrt{3} {/}2$ $\Rightarrow \;\; \theta =30^{\circ} \;\; [ \because \cos 30^{\circ}=\sqrt{3} {/} 2]$

Collision between A and B ${m} \\times 9={mv}_{1}+2 {m} {v}_{2}$ (from momentum conservation) ${e}=1=\frac{{v}_{2}-{v}_{1}}{9}$ $\Rightarrow{v}_{2}=6 {m} / {sec}, {v}_{1}=-3 {m} / {sec}$ collision between B and $C$ $2m \times 6 = 4mv$ (from momentum conservation) $v = 3$ m/s

As we know the centre of mass of the quarter disc, $=(\;\frac{4 R}{3 \pi }, \;\frac{4 R}{3 \pi })$ Since $R=a$ $\therefore$ The centre of mass of the quarter disc is $(\;\frac{4 a}{3 \pi }, \;\frac{4 a}{3 \pi })$. Hence, the value of $x$ is 4 .

Just after collision From momentum conservation, ${P}_{{i}}^{0}={P}_{{f}}$ ${\mu}={Mv} \.......$ (i) From angular momentum conservation about ${O}$, $m u \.\frac{L}{2}=\frac{M L^{2}}{12} \omega$ $\Rightarrow\omega =\frac{6 {\mu}}{{ML}}$ ........(ii) From e $=\frac{\text{R.V.S }}{\text{R.V.A }}$ $1={{V}+\frac{\omega {L}}{2}}/{{u}}$ ${v}+\frac{\omega {L}}{2}={u}$ ${v}+\frac{3 {\mu}}{{M}}={u}$ $\frac{{\mu}}{{M}}+\frac{3 {\mu}}{{M}}={u}$ $\frac{4 {\mu}}{{M}}={u}$ $\frac{{m}}{{M}}=\frac{1}{4}$ ${X}=4$

Let $v$ is the speed of the third block $C$. The velocity of centre of mass of $A$ and $B$ is $v_{C M}=\;\frac{v}{2}$ The spring is compressed maximum by $x$ distance. Using the law of conservation of energy. $\;\frac{1}{2} m v^{2}=\;\frac{1}{2} m(\;\frac{v}{2})^{2}+\;\frac{1}{2} m(\;\frac{v}{2})^{2}+\;\frac{1}{2} k x^{2}$ $\Rightarrow \;\; \;\frac{1}{4} m v^{2}=\;\frac{1}{2} k x^{2} \Rightarrow x=\sqrt{\;\frac{m v^{2}}{2 k}}$ $\Rightarrow \;\; x=v \sqrt{\;\frac{m}{2 k}}$

${K}_{2}=4 {K}_{1}$ $\frac{1}{2} {mv}_{2}^{2}=4 \frac{1}{2} {mv}_{1}^{2}$ ${v}_{2}=2 {v}_{1}$ ${P}={mv}$ ${P}_{2}={mv}_{2}=2 {mv}_{1}$ ${P}_{1}={mv}_{1}$ $\%$ change $=\frac{\Delta{P}}{{P}_{1}} \\times 100$ $=\frac{2 {mv}_{1}-{mv}_{1}}{{mv}_{1}} \\times 100=100 \%$

The mass $m_{1}$ is moving with speed $u_{1}$ initially and mass $m_{2}$ is at rest. After the collision, the mass $m_{1}$ and $m_{2}$ move with speed $v$ in opposite directions.Using the law of conservation of linear momentum, $m_{1} U_{1}+m_{2} U_{2}=m_{1} v_{1}+m_{2} v_{2}$ $\Rightarrow \;\; m_{1} u_{1}+m_{2}(0)=m_{1}(-v)+m_{2} v$ $\Rightarrow \;\; m_{1} u_{1}=(-m_{1}+m_{2}) v$...(i) Since, the collision is elastic because they move with same speed after the collision. Hence, coefficient of restitution, $e=1$ $\therefore \;\; {e}=\;\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \Rightarrow 1=\;\frac{v-(-v)}{u_{1}-0}$ $u_{1}=2 v$...(ii) Putting the above value in Eq. (i), we get $m_{7}(2 v)=(-m_{1}+m_{2}) v$ $\Rightarrow \;\; 3 m_{1}=m_{2} \Rightarrow \;\frac{m_{2}}{m_{1}}=\;\frac{3}{1}$ $\Rightarrow \;\; m_{7}: m_{1}=3: 1$

${p}_{{i}}={p}_{{f}}$ $2 \\times 4=2 \\times 1+{m}_{2} \\times {v}_{2}$ ${m}_{2} {v}_{2}=6$ .......(i) by coefficient of restitution $1=\frac{{v}_{2}-1}{4} \Rightarrow{v}_{2}=5 {m} / {s}$ by (i) ${m}_{2} \\times 5=6$ ${m}_{2}=1.2 {kg}$ ${v}_{{cm}}=\frac{{m}_{1} {v}_{1}+{m}_{2} {v}_{2}}{{m}_{1}+{m}_{2}}$ ${v}_{{cm}}=\frac{2 \\times 1+1.2 \\times 5}{2+1.2}$ $=\frac{8}{3.2}=\frac{25}{10}$ ${x}=25$

Let $v_{1}$ and $v_{2}$ are the speed of $P$ and $Q$ after collision. By using law of conservation of mementum, $m_{1} u_{1}+m_{2} u_{2} \;\; =m_{1} v_{1}+m_{2} v_{2}$ $M u+m \cdot 0 \;\; =M v_{1}+m v_{2}$ $\Rightarrow \;\; \;\; \;\frac{M(u-v_{1})}{m} \;\; =v_{2}$ and by using law of conservation of energy, $\;\;\frac{1}{2} m_{1} u_{1}^{2}+\;\frac{1}{2} m_{2} u_{2}^{2} \;\; =\;\frac{1}{2} m_{1} v_{1}^{2}+\;\frac{1}{2} m_{2} v_{2}^{2}$ $\;\Rightarrow \;M u^{2}+0 \;\; =M v_{1}^{2}+m v_{2}^{2}$ $\Rightarrow \;M(u^{2}-v_{1}^{2}) \;\; =m v_{2}^{2}$ $\Rightarrow \;M(u-v_{1})(u+v_{1}) / m \;\; =v_{2}^{2}$ $\;\frac{1}{2} m_{1} u_{1}^{2}+\;\frac{1}{2} m_{2} u_{2}^{2} \;\; =\;\frac{1}{2} m_{1} v_{1}^{2}$ $M u^{2}+0 \;\; =M v_{1}^{2}+m v^{2}$ $\Rightarrow \;\; M(u^{2}-v_{1}^{2}) \;\; =m v_{2}^{2}$ $\Rightarrow \;\; M(u-v_{1})(u+v_{1}) / m \;\; =v_{2}^{2}$ $\;\text{ Substituting the value of }\; M \;\frac{(u-v_{1})}{m} \;\text{ from Eq. (i) \in }\;\;$ $\;\text{ Eq. (ii), we get }\;$ $v_{2}(u+v_{1}) \;\; =v_{2}^{2}$ $u+v_{1} \;\; =v_{2}$ $\Rightarrow \;\; M\;≫ m$ $v_{1} \;\; =u$ $v_{2} \;\; =2 u$ Eq. (ii), we get $v_{2}(u+v_{1}) \;\; =v_{2}^{2}$ $u+v_{1} \;\; =v_{2}$ $M > \; > m$ $v_{1} \;\; =u$ $v_{2} \;\; =2 u$ Hence, option (c) is the correct.

Given, $m_{A}=1 {kg}, m_{B}=2 {kg},( {KE})_{A}:( {KE})_{B}=A: 1$ Linear momentum of $A$ and $B$ are equal. $\Rightarrow \;\; p_{A}=p_{B}$ $\because$ Kinetic energy $( {KE})=p^{2} / 2 m$ $\therefore \;\; \;\frac{ {KE}_{A}}{ {KE}_{B}}=\;\frac{m_{B}}{m_{A}}=\;\frac{2}{1}=\;\frac{A}{1} \Rightarrow A=2$