Centre Of Mass And Collision Practice Questions

$E_{\text{initial}}=1/2m(2v)^{2+}1/2 2m(v)^2=3mv^2$ $E_{\text{final}}=1/2 3m(4/9v^{2+}4/9v^2)=4/3mv^2$∴Fractional loss=${2-4/3}/3=5/9=56$

$Z_0=h-h/4={3h}/4$

Loss of energy is maximum when collision is inelastic as in an inelastic collision there will be maximum deformation.KE in COM frame is $1/2(\frac{Mm}{M+m})V^2_{\text{rel}}$ $KE_i1/21/2(\frac{Mm}{M+m})V^2\;\;KE_f=0 (\because V_{rel}=0)$Hence loss in energy is $1/2(\frac{Mm}{M+m})V^2$ $\Rightarrow f=M/{M+m}$

In completely inelastic collision,$m_1 v_1+m_2 v_2=m_1 v+m_2 v$after collision both particle have common velocity $v$, so all energy is not lost$\therefore$ Statement $-1$ is trueThe principle of conservation of momentum applicable for all kinds of collisions.So statement $-2$ is also true.Statement - 2 explains statement - 1 correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body.

Motion is uniform as graph is straight line. In ${\text{x}}$-t graph, velocity is $\;\frac{x}{t}$ and the slope of the graph tells the direction of the velocity. Here because of impulse the direction of velocity or slope of the graph changes.During each collisionInitial velocity $v_1=\;\frac{2}{2}=1 {\text{ms}}^{-1}$Final velocity $v_2=-\;\frac{2}{2}=-1 {\text{ms}}^{-1}$Impulse $=$ Change in momentum$=m |v_2-v_1{{}/{}|} =0.4 \times 2=0.8 {\text{Ns}}$

Let $m=0.50 {\text{kg}}$ and $M=1.0 {\text{kg}}$ Initial kinetic energy of the system when $1 {\text{kg}}$ mass is at rest, $K . E_i=\;\frac{1}{2} m u^{2+}\;\frac{1}{2} M(0)^2$ $=\;\frac{1}{2} \times 0.5 \times 2 \times 2+0=1 J$ For collision, applying conservation of linear momentum $m \times u=(m+M) \times v$ $\therefore 0.5 \times 2=(0.5+1) \times v \Rightarrow v=\;\frac{2}{3} m / s$ Final kinetic energy of the system is $K . E_f=\;\frac{1}{2}(m+M) v^2$ $=\;\frac{1}{2}(0.5+1) \times \;\frac{2}{3} \times \;\frac{2}{3}=\;\frac{1}{3} J$ $\therefore$ Energy loss during collision$= (1-\;\frac{1}{3}) J=0.67 J$

Given The linear mass density $\lambda =k (\;\frac{x}{L}) ^n$ $x_{C M}=\;\frac{∫_{0}^{L} x d m}{∫_{0}^{L} d m}$ $=\;\;\frac{∫_{0}^{L} x(\lambda \text{dx})}{∫_{0}^{L} \lambda \text{dx}}$ $=\;\;\frac{∫_{0}^{L} k (\frac{x}{L}) ^n \cdot x \text{dx}}{∫_{0}^{L} k (\;\frac{x}{L}) ^n \text{dx}}$ =\;\;{k \[\frac{x^{n+2}}{(n+2) L^n}]$ 0^L}/{[;\frac{k x^{n+1}}{(n+1) L^n}]$ 0^L}$$=;;\frac{L(n+1)}{n+2}$For$n=0, x{C M}=;\frac{L}{2} ; n=1$,$x{C M}=;\frac{2 L}{3} ; n=2, x_{C M}=;\frac{3 L}{4} ; . . .$From here you can see only option (A) can be the right answer.

Let the mass per unit area be $\sigma$. Then the mass of the complete disc=\sigma \[π(2 R)^2]$ =4 \pi \sigma R^2$The mass of the removed disc$=\sigma (\pi R^2) =\pi \sigma R^2$So mass of the remaining disc$=4 \pi \sigma R^{2-}\pi \sigma R^2=3 \pi \sigma R^2$Let center of mass of$3 \pi \sigma R^2$mass is at${\text{x}}$distance from origin 0$\therefore ;\frac{3 \pi R^2 \sigma . x+\pi R^2 \sigma . R}{4 \pi R^2 \sigma }=0$As center of mass of full disc is at Origin.$\therefore x=-;\frac{R}{3}$According to the question,$x=\alpha R$$\therefore \alpha =-;\frac{1}{3}$$\Rightarrow |\alpha |=;\frac{1}{3}$

We know, Force $\times$ time $=$ Impulse $=$ Change in momentum$\; \therefore F \times t=m(v-u)$ $\;\Rightarrow F=\;\frac{m(v-u)}{t}=\;\frac{0.15(0-20)}{0.1}=30 N$

Here linear momentum is conserved as no external force is acting on the bomb.Let the velocity and mass of $4 {\text{kg}}$ piece be $v_1$ and $m_1$ and that of $12 {\text{kg}}$ piece be $v_2$ and $m_2$Applying conservation of linear momentum$\;0=m_2 v_2-m_1 v_1$ $\;\Rightarrow v_1=\;\frac{12 \times 14}{4}=12 m s^{-1}$ $\; \therefore K . E_1=\;\frac{1}{2} m_1 v_1^2=\;\frac{1}{2} \times 4 \times 144=288 J$

$0=\;\frac{m_1 (-x_1) +m_2 x_2}{m_1+m_2} \Rightarrow m_1 x_1=m_2 x_2$ Finally,mass $m_1$ moved towards the center a distance $d$ so the distance of mass $m_1$ from the origin is $x_1-d$, and now let mass $m_2$ need to move $d^{'}$ to keep the center at the origin. $\therefore 0=\;\frac{m_1 (d-x_1) +m_2 (x_2-d^{'}) }{m_1+m_2}$ $\Rightarrow 0=m_1 d-m_1 x_1+m_2 x_2-m_2 d^{'}$ $\Rightarrow d^{'}=\;\frac{m_1}{m_0} d$

This is a case of translation motion without rotation, the force $\vec{F}$ has to be applied at center of mass.i.e. the point ${ }^{'} P^{'}$ has to be at the centre of mass$\;y=\;\frac{m_1 y_1+m_2 y_2}{m_1+m_2}$ $\;=\;\frac{m \times 2 ℓ+2 m \times ℓ}{3 m}$ $\;=\;\frac{4 ℓ}{3}$

Assume speed of second mass $=v_1$As momentum is conserved,In $x$-direction, $m v=m v_1 \cos \theta$In $y$-direction, $\;{m v}/{{\sqrt{3}}}=m v_1 sin \;\theta$Squaring and adding eqns.(1) and (2)$\; (m v_1 \cos \theta ) ^{2+} (m v_1 sin \;\theta ) ^2=(m v)^{2+} (\;{m v}/{{\sqrt{3}}}) ^2$ $\;\Rightarrow v_1^2=v^{2+}\;{v^2}/{{\sqrt{3}}}$ $\;\Rightarrow v_1=\;{2}/{{\sqrt{3}}} v$

Elastic energy stored in the spring $=\;\frac{1}{2} k L^2$And kinetic energy of the block $=\;\frac{1}{2} M v^2$ $\; \therefore \;\frac{1}{2} M v^2=\;\frac{1}{2} k L^2$ $\;\Rightarrow v=\sqrt{\;\frac{k}{M}} \cdot L$∴ Momentum$=M \times v=M \times \sqrt{\;\frac{k}{M}} \cdot L={\sqrt{k M}} \cdot L$

The center of mass does not shift as no external force is applied horizontally. So the center of mass of the system continues its original path. It is only the internal forces which comes into play while breaking

Assume the man can fire $n$ bullets in one second.$\therefore$ change in momentum per second $=n \times m v=F$[ $m=$ mass of bullet, $v=$ velocity, $F=$ force) $]$ $\therefore n=\;\frac{F}{m v}=\;\frac{144 \times 1000}{40 \times 1200}=3$

Kinetic energy of a system of particle is zero this is possible only when the speed of each particles is zero. And if speed of each particle is zero, the linear momentum of the system of particle has to be zero.Also the linear momentum of the system may be zero even when the particles are moving in different direction. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero.$\therefore A$ does not implies $B$ but $B$ implies $A$.

The velocity of center of mass of two particle system is $v_c=\;\frac{m_1 v_1+m_2 v_2}{m_1+m_2}$ $=\;\frac{m(2 v)+m(-v)}{m+m}$ $=\;\frac{v}{2}$