Center of Mass and Collision

Given, volume of block $A=0.25 \mathrm{~m}^3$

Let the density of iron be $\rho \mathrm{kg} / \mathrm{m}^3$.

Thus, mass of iron block, $A=0.25 \rho\mathrm{~kg}$

When block $A$ is attached to the spring elongation produced $(x)=0.2 \mathrm{~m}$.

Thus, force on spring of spring constant $k$ is

$F=-k x$

($-$ve sign denotes force is in opposite direction of elongation)

$\begin{aligned} m g & =-k(0.2) \\ 0.25 \rho g & =-0.2 k \Rightarrow k=1.25 \rho g \mathrm{~N} / \mathrm{m} \end{aligned}$

Now, the same spring is attached to another block $B$.

Volume of block $B=0.75 \mathrm{~m}^3$

Mass of block $B=0.75 \rho \mathrm{~kg}$

FBD of the spring system on inclined surface is given as

$\begin{aligned} & \text { Force on block } B, F=-(k x) \\ & M g \sin \theta=-k x \\ & 0.75 \rho g \sin 30^{\circ}=-(-1.25 \rho g) x \\ & \frac{0.75}{1.25} \sin 30^{\circ}=x \Rightarrow \frac{0.75}{1.25} \times \frac{1}{2}=x \\ & \Rightarrow \quad x=0.3 \mathrm{~m} \\ \end{aligned}$

Thus, the distance of block from the top = original length of spring + elongation = 1 m + 0.3 m = 1.3 m

Given, mass of ball $m=0.5 \mathrm{~kg}$

Initial velocity, $u=10 \mathrm{~ms}^{-1}$

Change in momentum, $\Delta p=8 \mathrm{~kg}-\mathrm{ms}^{-1}$

According to given situation change in momentum of ball

$\begin{aligned} & & \Delta p & =p_f-p_i \\ \Rightarrow & & 8 & =m v-m(-u) \\ \Rightarrow & & 8 & =m(v+u) \\ \Rightarrow & & 8 & =0.5(v+10) \\ \Rightarrow & & 16 & =v+10 \\ \Rightarrow & & v & =6 \mathrm{~ms}^{-1} \end{aligned}$

distribution of masses is given as $m\left(\frac{1}{3}\right)^N \frac{1}{N}$

Which are placed at $x=N$

when, $N=2,3,4 \ldots \ldots . . \infty$

Centre of mass of this discrete system is given as

$X=\frac{\Sigma m_i x_i}{\Sigma m_i}$

where, $\Sigma m_i$ is the total mass, which is given as $M$.

Thus,

$\begin{aligned} & X=\frac{\sum m_i x_i}{M} \\ &=\frac{1}{M} \sum_{n=2}^{\infty} m\left(\frac{1}{3}\right)^N \frac{1}{N} \times N \\ &=\frac{m}{M} \sum_{n=2}^{\infty}\left(\frac{1}{3}\right)^N \\ &=\frac{m}{M}\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4} \ldots \ldots \frac{1}{3^{\infty}}\right) \\ &=\frac{m}{M} \times \frac{1}{9}\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3} \ldots . . .\right) \\ &=\frac{m}{M} \times \frac{1}{9}\left(\frac{1}{1-\frac{1}{3}}\right) \\ &(\because S_n=\frac{a}{1-r} \text { when } r<1 \text { in G.P series) } \\ &=\frac{m}{M} \times \frac{1}{9} \times \frac{3}{2}=\frac{m}{6 M} \end{aligned}$

According to the law of conservation of linear momentum at the highest point,

$mv\cos \theta = {m \over 2}(v\cos \theta ) + {m \over 2}v'$

$ \Rightarrow mv\cos \theta - {m \over 2}v\cos \theta = {m \over 2}v'$

$ \Rightarrow v' = 2\left( {v\cos \theta - {{v\cos \theta } \over 2}} \right) = v\cos \theta $

According to given diagram,

Mass of ball, m = 3 kg

Speed of ball, v = 100 ms$^{-1}$

Time of contact, t = 0.2 s

As we know that,

$F=\frac{\Delta p}{\Delta t}$

where, F = force

$\Delta p$ = change in momentum

$\Delta t$ = change in time

$\therefore \quad F=\frac{2mv\cos30^\circ}{0.2}$

$=\frac{2\times3\times100\times\frac{\sqrt3}{2}}{0.2}=1500\sqrt3$ N

Given, mass of ball 1, $m_1=2 \mathrm{~kg}$,

Mass of ball 2, $m_2=3 \mathrm{~kg}$

$\begin{aligned} & u_1=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s} \\ & u_2=0 \end{aligned}$

After collision, the two balls will stick together and their combined velocity be $v$.

Now, by using law of conservation of momentum,

$\begin{aligned} m_1 u_1+m_2 u_2 & =\left(m_1+m_2\right) v \\ \Rightarrow \quad v & =\frac{2 \times 10+3 \times 0}{5}=4 \mathrm{~ms}^{-1} \end{aligned}$

$\therefore$ Energy loss $=$ Energy before collision $\left(E_2\right)$ $-$ Energy after collision $\left(E_1\right)$

$\begin{aligned} & =\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\frac{1}{2}\left(m_1+m_2\right) v^2 \\ & =\left(\frac{1}{2} \times 2 \times 10^2+\frac{1}{2} \times 3 \times 0^2\right)-\frac{1}{2}(2+3) 4^2 \\ & =\frac{1}{2} \times 2 \times 100-\frac{1}{2} \times 5 \times 16 \\ & =100-40=60 \mathrm{~J} \end{aligned}$

According to graph

Since, $F=$ rate of change of momentum $=d p / d t$

$\begin{aligned} \Rightarrow \quad \int_{p_1}^{p_2} d p & =\int_0^T F d t=\text { Area under graph } \\ & =\frac{1}{2} \times \frac{T}{4} \times F_0+\frac{2 T}{4} \times F_0+\frac{1}{2} \times \frac{T}{4} \times F_0 \\ & =\frac{F_0 T}{8}+\frac{2 F_0 T}{4}+\frac{F_0 T}{8}=\frac{6 F_0 T}{8} \\ \Rightarrow \quad p_2-p_1 & =\frac{3 F_0 T}{4} \Rightarrow m v-0=\frac{3 F_0 T}{4} \\ \Rightarrow \quad F_0 & =\frac{4 m v}{3 T} \end{aligned}$

We know that,

$\mathbf{r}=\frac{m_1 \mathbf{r}_1+m_2 \mathbf{r}_2+m_3 \mathbf{r}_3+\ldots+m_n \mathbf{r}_n}{M}$ ...... (i)

Let origin be the centre of mass of the body which is represented by $C$.

i.e. $r = 0$

Let origin be the centre of mass of the body which is represented by $C$.

i.e. $\mathbf{r}=0$

$\therefore$ From Eq. (i), we get

$m_1 \mathbf{r}_1+m_2 \mathbf{r}_2+m_3 \mathbf{r}_3+\ldots m_n \mathbf{r}_n=0$ .... (ii)

Net moment of all particles in the system about centre of mass $\mathrm{C}$.

$\begin{array}{rlr} \boldsymbol{\tau}_C & =m_1 g \mathbf{r}_1+m_2 g \mathbf{r}_2+m_3 g \mathbf{r}_3+\ldots m_n g \mathbf{r}_n \\\\ & =g\left[m_1 \mathbf{r}_1+m_2 \mathbf{r}_2+m_3 \mathbf{r}_3+\ldots+m_n \mathbf{r}_n\right] \\\\ & =g \times 0 \quad \text { [from Eq. (ii) }] \\ & =0 & \end{array}$

Hence, sum of moments of all the particles in a system about its centre of mass is always zero.

The given situation is shown in the following figure.

By the law of conservation of linear momentum, we have Total momentum before collision = Total momentum after collision

$ \begin{aligned} & m_1(u)+m_2(0)=m_1\left(\frac{2 u}{3}\right)+m_2(v) \\ & \Rightarrow \quad m_1 u+0=\frac{2}{3} m_1 u+m_2 v \end{aligned} $

$ \begin{align*} & \Rightarrow \quad m_1 u-\frac{2}{3} m_1 u=m_2 v \Rightarrow \frac{1}{3} m_1 u=m_2 v \\ & \Rightarrow \quad \frac{m_1}{m_2}=\frac{3 v}{u} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Since, the collision is perfectly elastic, then coefficient of restitution $=1$

$ \begin{aligned} e & =1 \\ \left|\frac{\mathbf{v}_2-\mathbf{v}_1}{\mathbf{u}_1-\mathbf{u}_2}\right| & =1 \Rightarrow \frac{v-\frac{2 u}{3}}{u-0}=1 \\ \Rightarrow \quad \frac{v-\frac{2 u}{3}}{u} & =1 \Rightarrow v=\frac{5 u}{3} \end{aligned} $

∴ From Eq. (i), we get

$ \frac{m_1}{m_2}=\frac{3 \times 5 u / 3}{u}=5 $

The given situation is shown in the following figure

Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$.

Hence, according to conservation of energy,

$ \begin{array}{rlrl} & & \frac{1}{2} M v_2^2 & =M g h \\ \Rightarrow & v_2 & =\sqrt{2 g h} \end{array} $

Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$

$ \begin{aligned} \Rightarrow \quad v_2 & =\sqrt{2 \times 10 \times 20 \times 10^{-2}} \\ & =2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Now, conservation of momentum gives

$ \begin{aligned} m u & =m v_1+M v_2 \\ \frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\ \frac{25}{4} & =\frac{25}{1000} \times v_1+2 \\ \Rightarrow \quad v_1 & =170 \mathrm{~ms}^{-1} \end{aligned} $

Let mass per unit area of the $\operatorname{disc}=m$

Total mass of disc, $M=\pi R^2 m$

Mass of the scooped out portion of the disc,

$ M^{\prime}=\pi r^2 m $

For small objects, centre of gravity and centre of mass are at same position.

We take the centre $O$ of the disc as the origin. The masses $M$ and $M^{\prime}$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively. The mass $M^{\prime}$ of the removed portion is taken negative.

The $x$-coordinate of the centre of mass of the remaining portion of the disc will be

$ \begin{aligned} & x_{\mathrm{CM}}=\frac{M x_1-M^{\prime} x_2}{M-M^{\prime}}=\frac{M \times 0-M^{\prime} \times 3}{\pi R^2 m-\pi r^2 m} \\ &=\frac{-3 M^{\prime}}{\pi m\left(R^2-r^2\right)} \\ &=\frac{-3 \pi r^2 m}{\pi m\left(R^2-r^2\right)} \\ &=-\frac{-3 r^2}{R^2-r^2}=\frac{-3 \times 3^2}{6^2-3^2}=-1 \mathrm{~cm} \\ & \Rightarrow \quad x_{\mathrm{CM}}=-1 \mathrm{~cm} \end{aligned} $

Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be 1 cm left side.