In Fig. 9-83, block 1 slides along an xaxis on a frictionless floor with a speed of 0.75 m/s. When it reaches stationary block 2, the two blocks undergo an elastic collision. The following table gives the mass and length of the (uniform) blocks and also the locations of their centers at time. Where is the center of mass of the two-block system located (a) at t=0, (b) when the two blocks first touch, and (c) att=4.0 s?

i) Mass of block 1 is $M_1=0.25\mathrm{kg}$.

ii) Mass of block 2 is $M_2=0.50\mathrm{kg}$.

iii) Velocity of block 1 before collision is $v_i=0.75\mathrm{m}/\mathrm{s}$.

iv) Velocity of block 2 before collision is $v_2=0.75\mathrm{m}/\mathrm{s}$.

v) Length of block 1 is 0.05m.

vi) Length of block 2 is 0.06m.

vii) Initial center of mass for block 1 is $x_{com,inital}=-1.50\mathrm{m}$.

viii) Initial center of mass for block 2 is 0m.

From the positions of blocks, find the position of their center of masses using the corresponding formula. Now, find the positions of the blocks after the two blocks touch each other. Using the same formula, find the position of the center of mass in this case. Then, find the velocity of center of mass which can be used to find the position center of mass at t=4.0 sec.

Formulae are as follow:

i)$x_{com}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

ii)$V_{com}=\frac{\left(v_1{m}_1+v_2{m}_2\right)}{\left(m_1+m_2\right)}$

Here, $m_1$, $m_2$are masses, $x_1$, $x_2$are points where m₁, m₂ are present,v₁, v₂ are velocities of $m_1$, $m_2$, $v_{com}$is velocity of centre of mass and $x_{cm}$is center of mass.

Centre of mass when t=0 is,

$x_{com}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

$x_{com}=\frac{\left(0.25\right)\left(-1.5\right)+\left(0.5\right)\left(0\right)}{0.25+0.5}$

$x_{com}=-0.50\mathrm{m}$

Hence, center of mass at t=0 is -0.50 m.

Center of each block is at half of the total length of each block. Left edge of block 2 is at zero. When block 1 touches block 2 , the right edge of block 2 will be at the same position of the left edge of block 1 which is at$\left(0-\left(\frac{5}{2}\right)\right)=-2.5\mathrm{cm}$ and center point of block 1 will be at$-2.5-3=-5.5\mathrm{cm}$

$x_{com}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

$x_{com}=\frac{\left(0.25\right)\left(-5.5\right)+\left(0.5\right)\left(0\right)}{0.25+0.5}$

$x_{com}=-1.83\approx -1.8c\mathrm{m}$

Hence, centre of mass when the two blocks touch is -1.8 cm.

In absence of external force center of mass moves with constant velocity. And velocity of center of mass is as follows:

$v_{com}=\frac{\left(v_1{m}_1+v_2{m}_2\right)}{\left(m_1+m_2\right)}$

$v_{com}=\frac{\left(0.25\right)\left(0.75\right)+\left(0.5\right)\left(0\right)}{0.25+.5}$

$v_{com}=0.25\frac{\mathrm{m}}{\mathrm{s}}$

So, center of mass at t=4 sec is as follows:

$$\begin{gathered} x_{com}=x_{com,inital}+v_{com}t \\ {\mathrm{x}}_{\mathrm{com}}=-0.5+\left(0.25\times 4\right) \\ {\mathrm{x}}_{\mathrm{com}}=0.50\mathrm{m} \end{gathered}$$

Hence, centre of mass at t=4.0 sec is 0.50m.

Therefore, the position and velocity of the center of mass of bodies can be found from positions and velocities of bodies using corresponding formulae.