14.4 Laws of motion and system of particles (check your understanding  (Page 5/2)

The particle is at rest before breaking into parts due to internal force. It means that the COM of the parts should remain at the initial position. In other words, its velocity should be zero even after breaking. Now, the speed of the COM is :

$$\begin{array}{l}{\mathbf{v}}_{\mathrm{COM}}=\frac{m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2}{m_1+m_2}=0\end{array}$$

$$\begin{array}{l}\Rightarrow m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2=0\end{array}$$

$$\begin{array}{l}{\mathbf{v}}_1=-\left(\frac{m_2}{m_1}\right){\mathbf{v}}_2\end{array}$$

Clearly, the velocities of the parts of the sphere should move in opposite directions. As the parts are unequal, they should move with different speeds as well.

Hence, option (d) is correct.

The acceleration of the system of two particles is given by :

$$\begin{array}{l}{\mathbf{a}}_{\mathrm{COM}}=\frac{m_1{\mathbf{a}}_1+m_2{\mathbf{a}}_2}{m_1+m_2}\\ {\mathbf{a}}_{\mathrm{COM}}=\frac{0.1x0+0.2x\mathbf{a}}{0.3}=\frac{2\mathbf{a}}{3}\end{array}$$

Hence, option (b) is correct.

Sphere is composed by a system of particle. As the density of sphere is uniform, its COM coincides with the center of sphere. Now, the acceleration of the "center of mass" is dependent on the net external force on the system. It does not depend on the point of application. Recall that we explained that COM represents a point where external force "appears" to apply - not the point where net external force acts.

Hence, option (d) is correct.

In the beginning when the blocks are released from rest, the COM of the two blocks system is also at rest. Once the blocks are released, they are acted upon by force due to gravity. We consider tensions in the string as the internal force. It means that COM is accelerated downward. In order to find the displacement of the COM, we need to know its vertical acceleration. The acceleration of the COM is obtained by :

$$\begin{array}{l}{\mathbf{a}}_{\mathrm{COM}}=\frac{m_1{\mathbf{a}}_1+m_2{\mathbf{a}}_2}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow {\mathbf{a}}_{\mathrm{COM}}=\frac{{\mathbf{a}}_1+2{\mathbf{a}}_2}{3}\end{array}$$

Thus, we need to know the accelerations of individual block. As the blocks are connected with a taut string, the magnitude of acceleration of each block is same, but they are directed opposite to each other. From force analysis on individual block,

For first block :

$$\begin{array}{l}\Rightarrow \sum F_y=T-mg=ma\end{array}$$

For second block :

$$\begin{array}{l}\Rightarrow \sum F_y=2mg-T=2ma\end{array}$$

Solving two equations, we have :

$$\begin{array}{l}\Rightarrow a=\frac{g}{3}\end{array}$$

Putting in the equation of acceleration of COM in y - direction with appropriate sign (downward direction as positive):

$$\begin{array}{l}\Rightarrow a_{\mathrm{COM}}=\frac{a_1+2a_2}{3}=\frac{-\frac{g}{3}+\frac{2g}{3}}{3}\end{array}$$

$$\begin{array}{l}\Rightarrow a_{\mathrm{COM}}=\frac{g}{9}=\frac{10}{9}\end{array}$$

Now, let "y" be the displacement of COM in 1 second,

$$\begin{array}{l}\Rightarrow y=\frac{1}{2}a{t}^2\end{array}$$

$$\begin{array}{l}\Rightarrow y=\frac{1}{2}x\frac{10}{9}x1=0.51m\end{array}$$

Hence, option (a) is correct.