Centre Of Mass And Collision Practice Questions

v = $\sqrt{2gl(1-cos\theta _0)}$ $v_1$ = $\sqrt{2gl(1-cos\theta _1)}$ By momentum conservation m $\sqrt{2gl(1-cos\theta _0)}$ = $MV_m$ - m $\sqrt{2gl(1-cos\theta )}$ \Rightarrow m $\sqrt{2gl}$ {\sqrt{1-cosθ_0}$+$\sqrt{1-cosθ_1}} = MV_m$and e = 1 =${V_m+\sqrt{2gl(1-cosθ_1)}}/\sqrt{2gl(1-cosθ_0)}$$\sqrt{2gl}$(\sqrt{1-cos\theta _0}-\sqrt{1-cos\theta _1}$) = $V_m$ ... (1) m $\sqrt{2gl}$ (\sqrt{1-cosθ_0}-\sqrt{1-cosθ_1}$) = M$V_M$... (2) Dividing$(\sqrt{1-cosθ_0}+\sqrt{1-cosθ_1})/(\sqrt{1-cosθ_0}-\sqrt{1-cosθ_1})$=$M/m$By componendo divided$\frac{m-M}{m+M}$=$\sqrt{1-cosθ_1}/\sqrt{1-cosθ_0}$=$\frac{sin(θ_1/2)}{sin(θ_0/2)}$\Rightarrow$M/m$=$\frac{θ_0-θ_1}{θ_0+θ_1}$\Rightarrow M =$\frac{θ_0-θ_1}{θ_0+θ_1}$

$m_{1} {\vec{r}}_{1}+m_{2} {r_{\vec{2}}}=M ( {\vec{r}_{c m}} )$ $\frac{M}{4} ( \frac{3 a}{4} \vec{i}+ \frac{3 b}{4} \vec{j})+ \frac{3 M}{4} \vec{r}_{2}$ $=M ( \frac{a}{2} \vec{i}+ \frac{b}{2} \vec{j})$ $\frac{3 M}{4} {\vec{r}_{2}}= ( \frac{a}{2}-\ \frac{3 a}{16} ) \vec{i}+ ( \frac{b}{2}- \frac{3 b}{16} ) \vec{j}$ $= \frac{5 a}{16} \vec{i}+ \frac{5 b}{16} \vec{j}$ ${r_{2}}= \frac{5 a}{12} \vec{i}+ \frac{5 b}{12} \vec{j}$

Momentum conservation $2 \\times v=2 \\times \frac{v}{4} \times m^{2} 1$ $m v_{1}= \frac{3 v}{2}$ Elatic collision Energy conservation $\ \frac{1}{2} \\times 2 \\times v^{2}= \frac{1}{2} \\times 2 \\times ( \frac{v}{4}\ )^{2}+\ \frac{1}{2} m v_{1}^{2}$ $\ \frac{2 v^{2}}{16} \\times 15=\ \frac{\ (m v_{1} )^{2}}{m} \Rightarrow\frac{v^{2} 15}{8}=\ \frac{9 v^{2}}{4 \\times m}$ $m=1.2 k g$

Velocity of 1 kg block just before it collideswith 3kg block = $\sqrt{2gh}$ = $\sqrt{2000}$ m/s Applying momentum conversation just beforeand just after collision 1 × $\sqrt{2000}$ = 4v \Rightarrow v = $\sqrt{2000}/4$ m/s initial compression of spring 1.25 × $10^6 x_0$ = 30 \Rightarrow $x_0$ ~ 0 applying work energy theorem $W_g + W_{sp}$ = ΔKE \Rightarrow 40 × x + $1/2$ × 1.55 × $10^6(0^{2-}x^2)$ = 0 - $1/2 \times 4 \times v^2$ solving x ~ 4 cm

$x_{\text{cm}}$ = $\frac{2mL+2mL+{5mL}/2}{4m}$ = ${13}/8$ L $Y_{\text{cm}}$ = $\frac{2m\times L+m\times (L/2)+m\times 0}{4m}$ = ${5L}/8$

By applying the principle of conservation of linear momentum in the $x$ and $y$ directions:
Along the x-axis: $M(10\cos30^\circ) + 2M(5\cos45^\circ) = 2M(v_1\cos30^\circ) + M(v_2\cos45^\circ)$
$\Rightarrow 5\sqrt{3} + \frac{10}{\sqrt{2}} = v_1\sqrt{3} + \frac{v_2}{\sqrt{2}} \Rightarrow 1.732v_1 + 0.707v_2 = 15.73$
Along the y-axis: $M(-10\sin30^\circ) + 2M(5\sin45^\circ) = 2M(v_1\sin30^\circ) - M(v_2\sin45^\circ)$
$\Rightarrow -5 + \frac{10}{\sqrt{2}} = v_1 - \frac{v_2}{\sqrt{2}} \Rightarrow v_1 - 0.707v_2 = 2.07$
Solving the system by adding the equations: $2.732v_1 = 17.8 \Rightarrow v_1 \approx 6.5 \text{ m/s}$.
Substituting $v_1$ yields $0.707v_2 = 6.5 - 2.07 = 4.43 \Rightarrow v_2 \approx 6.3 \text{ m/s}$.

$k_i$ = $1/2 mv_0^2$ From linear momentum conservation $mv_0$ = (2m + M) $v_f$ \Rightarrow $v_f$ = {mv_0]/{2m+M} $k_i/k_f$ = 6 \Rightarrow $\frac{1/2mv_0^2}{1/2(2m+M)(\frac{mv_0}{2m+M})^2}$ = 6 \Rightarrow ${2m+M}/m$ = 6 \Rightarrow $M/m$ = 4

$V_{1}+V_{2}=0.70$ $m_{1} V_{1}=-m_{2} V_{2}$ $20\ (0.70-V_{2}\ )=-50 \ \times V_{2}$ $14-20 V_{2}=50 V_{2}$ -30 V_{2}=14 \ $V_{2}=-0.47 \ {m}$ / ${s}$

$\frac{dp}{dt}$ = F = kt $∫^{3P}_{P}$ dP = $∫^{T}_{0}$ kt dt 2p = ${KT^2}/2$ T = 2 $\sqrt{p/k}$

$F=2 \frac{d N}{d t} \times m v$ $=2 \times 10^{22} \times 10^{-26}\times 10^{4}=2 N$Pressure $=2 N m^{-2}$

$m_{1} \vec{v}_{1}+m_{2} {\vec{v} _{2}}=m_{1} {\vec{v}_{3}}+m_{2} {\vec{v}_{4}}$ $2 m_{2}=m_{1}$ and Let $m_{2}=m$ $2 m v_{1}+m v_{2}=2 m (0.5 v_{1} )+m v_{4}$ $\Rightarrow v_{1}=v_{4}-v_{2}$

Time taken for the particles to collide, t = $d/V_{\text{rel}}$ = $\frac{100}{100}$ = 1 sec Speed of wood just before collision = gt = 10 m/s& speed of bullet just before collision v-gt = 100 - 10 = 90 m/s Now, conservation of linear momentum just before and after the collision - -(0.02) (1v) + (0.02) (9v) = (0.05)v \Rightarrow 150 = 5v ⇒ v = 30 Max height reached by the body h = $v^2/{2g}$ h = $\frac{30\times 30}{2\times 10}$ = 45 m ∴ Height above tower = 40 m

$\ \frac{1}{2} \\times \mu _{\ {red}} \times V_{\ {rel}}^{2}=m g h_{\max }$ $\frac{1}{2} \\times \frac{4 m}{5} \\times V^{2}=m g h_{\max }$ $\Rightarrow h_{\max }$ $=\ \frac{2 v^{2}}{5 g}$

$v^{2}-u^{2}=-2 g h$ $v^{2}=v^{2}-2 g h$ $p^{2}=A-B h$

Conservation of momentum: $m\vec{u} = m\vec{v}_1 + M\vec{v}_2$. Since $\vec{v}_1 \perp \vec{v}_2$, the vector diagram forms a right triangle: $m^2 u^2 = m^2 v_1^2 + M^2 v_2^2$.
Conservation of kinetic energy: $\frac{1}{2}m u^2 = \frac{1}{2}m v_1^2 + \frac{1}{2}M v_2^2 \implies M v_2^2 = m(u^2 - v_1^2)$.
Substituting $M v_2^2$ into the momentum equation: $m^2 u^2 = m^2 v_1^2 + M [m(u^2 - v_1^2)]$.
Expanding this: $m^2 u^2 = m^2 v_1^2 + M m u^2 - M m v_1^2$.
Rearranging: $m^2(u^2 - v_1^2) = Mm(u^2 - v_1^2)$.
Dividing both sides by $m(u^2 - v_1^2)$, we get $m = M$.

$P=F/A=\frac{\Delta (mv_1)}{\Delta t}=(2mv_1)/A(n/t)$ $\frac{2\times 3.32\times 10^{-27}\times (1000\sqrt2)}{2\times 10^{-4}}(10^{23})$ $=2.35\times 10^3N\text{/}m^2$

Momentum conservation$mv_0=m(v_1-v_2)$ $v_1=v_2=v_0$.............(1)$1/2m(v_1^{2+}v_2^2)-1/2mv_0^2=1/2\times 1/2mv_0^2$ $v_1^{2+}v_2^2=3/2v_0^2$.............(2)$(v_1-v_2)^2=v_0^2$ $v_1^{2+}v_2^{2-}2v_1v_2=v_0^2$.........(3)(2) - (3)$\Rightarrow 2v_1v_2=v_0^2/2$ $\Rightarrow 4v_1v_2=v_0^2$ $(v_1+v_2)=\sqrt{(v_1-v_2)^{2+}4v_1v_2}$Relative velocity after collision$=\sqrt{v_0^{2+}v_0^2}$ $=\sqrt{2v_0}$

Using conservation of linear moments, Along X-axis, $2 M V^{'} \cos \theta =M V \sin 30^{\circ}-M_{V} \sin 45^{\circ} . . . . . .$ (1) Along Y-axis, $2 {Mv}^{'} \sin \theta = {Mv} \cos 30^{\circ}+ {Mv} \cos 45^{\circ} . . . . . .$ (2) Dividing (2) by (1),we get, $\;\frac{\sin \theta }{\cos \theta }=\;\frac{\cos 30^{4}+\cos 45^{\circ}}{\sin 30^{\circ}-\sin 45^{\circ}}$ $=\;{\;\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}}/{\frac{1}{2}-\;\frac{1}{\sqrt{2}}}$ $\therefore \tan \theta =\;\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}$