Centre Of Mass And Collision Practice Questions

$1 \times {\text{u}}_1=-2+3 {\text{v}}\;\Rightarrow {\text{u}}_1=-2+3 {\text{v}}$ . . . (1)$1=\;{ {\text{v}}+2}/{ {\text{u}}_1}\;\Rightarrow {\text{v}}+2= {\text{u}}_1$ . . . (2)Solving (1) and (2)${\text{u}}_1=4 {\text{m}} / {\ s}$

Given,Mass of body $A=5 {\text{kg}}$Mass of body $B=8 {\text{kg}}$Momentum of body $B$ is twice that of body $A$,$\therefore P_B=2 P_A$We know,Kinetic Energy $(K)=\;\frac{P^2}{2 m}$ $\therefore \;\frac{K_A}{K_B}= (\;\frac{P_A}{P_B}) ^2 \times \;\frac{m_B}{m_A}$ $= (\;\frac{1}{2}) ^2 \times \;\frac{8}{5}$ $=\;\frac{1}{4} \times \;\frac{8}{5}$ $=\;\frac{2}{5}$

Taking the system as man and trolley and using conservation of linear momentum.$60 \times {\text{v}}=(60+120) \times 2$ $\Rightarrow {\text{v}}=6 {\text{m}} / {\ s}$

For a head on elastic collision$v_2=\;\frac{m u_1}{m+5 m}+\;\frac{m u_1}{m+5 m}$ $=\;\frac{2 u_1}{6} \;\text{ or }\; \;\frac{u_1}{3}$Initial kinetic energy of first mass $=\;\frac{1}{2} m u_1^2$Final kinetic energy of second mass$=\;\frac{1}{2} \times 5 m (\;\frac{u_1}{3}) ^2$ $=\;\frac{5}{9} (\;\frac{1}{2} m u_1^2)$ $\Rightarrow$ kinetic energy transferred $=55 \%$ of initial kinetic energy of first colliding mass

Change in momentum of one ball$=2 \times (0.05)(10) {kg} {m} / {s}$ $=1 {kg} {m} / {s}$ $\Rightarrow F_{a v g}=\;\frac{1}{\Delta t}=\;\frac{1}{0.005} {N}$ $=200 {N}$

Conserving momentum :$m(30 {i}^{∧})+m(40 {j}^{∧})+2 m(\vec{v})=\vec{0}$ $\Rightarrow \overline{v}=-15 {i}^{∧}-20 {j}^{∧}$ $\Rightarrow |v|=25 {\text{m}} / {\ s}$

${\text{K}} . {\text{E}}=\;{ {\text{P}}^2}/{2 {\text{m}}}$ ${\text{K}}_1=\;{ {\text{P}}_1^2}/{2(8)} ; {\text{K}}_2=\;{ {\text{P}}_2^2}/{2(2)}$ ${\text{K}}_1= {\text{K}}_2$ ${\text{So}}_0$ $4 {\text{P}}_2^2= {\text{P}}_1^2$ $\;{ {\text{P}}_1}/{ {\text{P}}_2}=2$

${\text{dm}}=\lambda \cdot {\text{dx}}=\lambda _0(1-\;{ {\text{x}}^2}/{ℓ^2})$ $X_{ {\text{cm}}}=\;{∫ {\text{xdm}}}/{∫ {\text{dm}}_{ℓ}}$ $=\;{\lambda _0 ∫_{0}^{ℓ}(x-\frac{x^3}{ℓ^2}) {\text{dx}}}/{∫_{0}^{ℓ} \lambda _0(1-\;\frac{x^2}{ℓ^2}) {\text{dx}}}=\;\frac{\;\frac{ℓ^2}{2}-\frac{ℓ^4}{4 ℓ^2}}{ℓ-\;\frac{ℓ^3}{3 ℓ^2}}=\;\frac{3 ℓ}{8}$

Let, initial momentum of body $(p_i) =p$ $\therefore$ Final momentum $(p_f) =p_i+20 \%$ of $p_i$ $=p+0.2 p$ $=1.2 p$We know,Kinetic energy $(E)=\;\frac{p^2}{2 m}$ $\therefore E_i=\;\frac{p^2}{2 m}$and $E_f=\;\frac{(1.2 p)^2}{2 m}=\;\frac{1.44 p^2}{2 m}$ $\therefore \%$ Change in kinetic energy$=\;\frac{E_f-E_i}{E_i} \times 100$ $=\;\frac{\;\frac{1.44 p^2}{2 m}-\frac{p^2}{2 m}}{\frac{p^2}{2 m}} \times 100$ $=\;\frac{\;\frac{p^2}{2 m}(1.44-1)}{\frac{p^2}{2 m}} \times 100=0.44 \times 100=44$

{ \vec{\text{P}}_{ {\text{i}}}=0.15 \times 12({ {\text{i}}^{\^}) { \vec{\text{P}}_{ {\text{f}}}=0.15 \times 12(-{ {\text{i}}^{\^}) |{\Delta \vec{\text{P}}|=3.6 {\text{kg}}- {\text{m}} / {\text{s}} $3.6= {\text{F}} \Delta {\text{t}}$ $3.6=100 \Delta {\text{t}}$ $\Delta {\text{t}}=0.036 {\sec}$

Impulse $=$ change in momentum ${\text{I}}=\Delta {\text{P}}$ ${\text{F}}_{\;\text{aug }\;}=\;{\Delta {\text{P}}}/{\Delta {\text{t}}}$ $\Delta {\text{t}}_1=3 \;\; \Delta {\text{t}}_2=5$ $\Delta {\text{P}}_1=\Delta {\text{P}}_2$ ${\text{I}}_1= {\text{I}}_2$ ${\text{F}}_{\;\text{avg }\;}$ in case (i) is more than (ii)

The position vector of the centre of mass ($\vec{r}_{cm}$) is given by:
$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$

1. Substitution

Substitute the given values $m_1 = 1$, $m_2 = 3$, $\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$, and $\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$:
$\vec{r}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$
$\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k}$

2. Magnitude Calculation

Calculate the magnitude of the resulting vector:
$|\vec{r}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + (1)^2}$
$|\vec{r}_{cm}| = \sqrt{4 + 1 + 1} = \sqrt{6}$

3. Comparison

The magnitude of the vector in Option A ($\hat{i} + 2\hat{j} + \hat{k}$) is:
$\sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$

Since the magnitudes are equal, the correct option is A.

For COM to remain unchanged,$m_1 x_1=m_2 x_2$ $\Rightarrow 10 \times 6=30 \times x_2$ $\Rightarrow x_2=2 {\text{cm}}$ towards $10 {\text{kg}}$ block.

$m=10 {\text{kg}}$ $\theta =45^{\circ}$ $y=x \tan \theta (1-\;\frac{x}{R})$ $\Rightarrow 10=20(1-\;\frac{20}{R})$ $\Rightarrow R=40$ $40=\;\frac{u^2}{10} \Rightarrow u=20$ $\Rightarrow T=\;{20 \times 20 \times \frac{1}{\sqrt{2}}}/{10}=\;\frac{4}{\sqrt{2}} s \Rightarrow t=2 s$at $t=2, \vec{v}=(10 \sqrt{2} \hat{i})+(10 \sqrt{2}-2 \times 10)$ $\Rightarrow \vec{p}=10[10 \sqrt{2} \hat{i}+(10 \sqrt{2}-20) \hat{j}]$ $=100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}$

$d_{c m}=3 \sin 45^{\circ}=\;\frac{3}{\sqrt{2}}$ $d_{c m}=\;\frac{2}{3} \times \;\frac{3}{\sqrt{2}}=\sqrt{2}=\sqrt{x}$ $x=2$

At maximum height it's velocity is zero. So momentum (mv) will be zero.

Impulse $=$ change in momentum$= {\ m}[ {\ v}-(- {\ v})]=2 {\ mv}$ $=2 \times 0.4 \times 15=12 {\ Ns}$

Given, mass of bullet $(m_{b})=10 {g}$ $=10 \times 10^{-3} {kg}$Initial speed of bullet is $v$ .Length of pendulum, I $=0.5 {m}$Mass of bob, $m=1 {kg}$Initial speed of bob, $u=0$Final speed of bullet, $v_{b}=100 {ms}^{-1}$Final speed of bob for making complete circle at bottom, $v^{'}=\sqrt{5 g l}$Acceleration due to gravity, $g=10 {ms}^{-2}$By using law of conservation of momentum $\;m_{b} u_{b}+m u \;\; =-m_{b} v_{b}+m v^{'}$ $\Rightarrow \;\;\frac{10}{1000} v+0 \;\; =\;\frac{-10}{1000} \times 100+1 \sqrt{5 g l}$ $\Rightarrow \;\;\frac{v}{100} \;\; =-1+\sqrt{5 \times 10 \times \;\frac{5}{10}}$ $\Rightarrow \;\;\frac{v}{100} \;\; =-1+5$ $\Rightarrow \;v \;\; =4 \times 100=400 {ms}^{-1}$

Given, mass of particle $A, m_{A}=4 {g}$ Mass of particle $B, m_{B}=16 {g}$ Kinetic energy of $A$ and $B$ is same i.e. $\;\; {KE}_{A}= {KE}_{B}$ As, kinetic energy $( {KE})=p^{2} / 2 m$ where, $p$ is momentum and $m$ is mass. $\therefore \;\; \;\frac{(p_{A})^{2}}{m_{A}}=\;\frac{(p_{B}^{2})}{m_{B}} \Rightarrow \;\frac{p_{A}^{2}}{4}=\;\frac{p_{B}^{2}}{16} \Rightarrow \;\frac{p_{A}}{p_{B}}=\;\frac{1}{2}$ $\because$ linear momentum is $n: 2$. $\therefore \;\; n=1$