A stone is dropped at${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$. A second stone, with twice the mass of the first, is dropped from the same point atrole="math" localid="1654342252844" ${}^{\mathbf{t}\mathbf{=}\mathbf{100}\mathbf{m}\mathbf{s}}$. (a) How far below the release point is the centre of mass of the two stones at${}^{\mathbf{t}\mathbf{=}\mathbf{300}\mathbf{ms}}$? (Neither stone has yet reached the ground.) (b) How fast is the centre of mass of the two stone systems moving at that time?

First stone is dropped at,${}^{\mathrm{t}=0\sec }$

Second stone is dropped at,${}^{\mathrm{t}=100\mathrm{ms}}$

Mass of second stone is twice the mass of first stone.

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system.

The expression for the coordinates of the center of mass are given as:

${\stackrel{\rightharpoonup }{r}}_{com}=\frac{1}{M}\sum _{i=1}^n{m}_i{r}_i$ … (i)

Here,${}^{\mathrm{M}}$is the total mass,${}^{{\mathrm{m}}_{\mathrm{i}}}$is the individual mass of ith particle and${}^{{\mathrm{r}}_{\mathrm{i}}}$is the coordinates of ith particle.

The expression for kinematic equations of motion in vertical direction are given as:

localid="1654344116642" $\mathrm{v}={\mathrm{v}}_0+\mathrm{gt}$ … (ii)

$\mathrm{y}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{gt}}^2$ … (iii)

Here,${}^{v_0}$is the initial velocity,${}^v$is the final velocity,${}^s$is the distance,${}^a$is the acceleration and${}^t$is the time.

The initial velocity for both the stones is zero.

Using equation (iii), the height of first stone is,

$$\begin{gathered} {\text{y}}_1={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{{\mathrm{gt}}^2}_1 \\ =0+\frac{1}{2}(9.8\mathrm{m}/{\mathrm{s}}^2){(0.3\mathrm{s})}^2=0.441\mathrm{m} \end{gathered}$$

Since the second stone is dropped after 100 ms, then${}^{{\mathrm{t}}_2=0.2\mathrm{s}}$

Using equation (iii), the height of second stone is,

$$\begin{gathered} {\text{y}}_2={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{{\mathrm{gt}}^2}_2 \\ =0+\frac{1}{2}(9.8\mathrm{m}/{\mathrm{s}}^2){(0.2\mathrm{s})}^2 \\ =0.196\mathrm{m} \end{gathered}$$

Now, using equation (i), the center of mass of two stones is,

$$\begin{gathered} {\mathrm{y}}_{\mathrm{co}\mathrm{m}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ =\frac{{\mathrm{m}}_1{\mathrm{y}}_1+2{\mathrm{m}}_1{\mathrm{y}}_2}{{\mathrm{m}}_1+2{\mathrm{m}}_1} \\ =\frac{({\mathrm{y}}_1+2{\mathrm{y}}_2)}{3} \end{gathered}$$

Substitute the values in the above expression.

role="math" localid="1654594305615" $$\begin{gathered} {\mathrm{y}}_{\mathrm{co}\mathrm{m}}=\frac{0.441\mathrm{m}+2\times 0.196\mathrm{m}}{3} \\ =0.277\mathrm{m} \\ \approx 0.28\mathrm{m} \end{gathered}$$

Thus, the center of mass of two stones at${}^{\mathrm{t}=300\mathrm{ms}}$is${}^{0.28\mathrm{ms}}$.

Using equation (ii), the velocity of first stone is,

$$\begin{gathered} {\mathrm{v}}_1={\mathrm{v}}_{01}+\mathrm{gt} \\ =0+(9.8\mathrm{m}/{\mathrm{s}}^2)(0.3\mathrm{s}) \\ =2.94\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (ii), the velocity of second stone is,

$$\begin{gathered} v_2=v_{02}+gt \\ =0+(9.8m/s^2)(0.2s) \\ =1.96m/s \end{gathered}$$

Now, by using center of mass formula we get

$$\begin{gathered} {\mathrm{v}}_{\mathrm{com}}=\frac{({\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2)}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ =\frac{{\mathrm{v}}_1+2{\mathrm{v}}_2}{3} \\ =\frac{(2.94)+2(1.96)}{3} \\ =2.287\mathrm{m}/\mathrm{s} \\ =2.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the center of mass is${}^{2.3\mathrm{m}/\mathrm{s}}$.