14.3 Laws of motion and system of particles  (Page 2/2)

Problem : Two spherical charged particles of 0.2 kg and 0.1 kg are placed on a horizontal smooth plane at a distance "x" apart. The particles are oppositely charged to a magnitude of 1μC. At what distance would they collide as measured from the initial position of the particle of 0.2 kg?

Solution : This is two particles system, which are acted by the electrostatic force of attraction. As the surface is smooth, we can neglect friction. We can also neglect weight of the particles and normal force as they are perpendicular to motion.

The important aspect to consider here is that electrostatic force of attraction between particles is internal force and, therefore, does not enter into our consideration. The position of collision, as we will see, is independent of the force of attraction and hence independent of the charge on the particles.

Now, as the external force on the particles is zero, the motion of the COM of the two particles system will not change. Since the particles were at rest in the beginning, the COM of the particle system will remain where it was in the beginning. At the time of the collision, when the particles have same position (approximately), the COM of particles is also the point of collision. It means that point of collision coincides with the COM. Now, COM of the particle system as measured from the first particle is :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{m_{2}x}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{0.1x}{0.3}=\frac{x}{3}\end{array}$$

Thus, charged particles collide at a distance "x/3" from the initial position of the particle of mass 0.2 kg.

Problem : Two particles of 0.2 kg and 0.3 kg, each of which moves with initial velocity of 5 m/s, collide at the origin of planar coordinate system. Find the velocity of their COM before and after the collision.

Solution : The components of velocities of COM are given as :

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMx}}=\frac{m_1{v}_{\mathrm{1x}}+m_2{v}_{\mathrm{2x}}}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMx}}=\frac{0.2X(-5\cos {37}^0)+0.3X0}{0.5}\end{array}$$

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMx}}=\frac{0.2X(-5X\frac{4}{5})}{0.5}=-1.6m/s\end{array}$$

and

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMy}}=\frac{m_1{v}_{\mathrm{1y}}+m_2{v}_{\mathrm{2y}}}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMy}}=\frac{0.2X(-5\sin {37}^0)+0.3X5}{0.5}\end{array}$$

$$\begin{array}{l}\Rightarrow v_{\mathrm{COMy}}=\frac{0.2X(-5X\frac{3}{5}))+1.5}{0.5}=1.8m/s\end{array}$$

Thus, velocity of COM is :

$$\begin{array}{l}\Rightarrow {\mathbf{v}}_{\mathrm{COM}}=-1.6\mathbf{i}+1.8\mathbf{j}\end{array}$$

Since there is no external force, the COM of the two particles move with same velocity before and after the collision.

Problem : A log of wood of length "L" and mass "M" is floating on the surface of still water. One end of the log touches the bank. A person of mass "m" standing at the far end of the log starts moving toward the bank. Determine the displacement of the log when the person reaches the nearer end of the log?

Solution : AIn this case, no external force is involved in the direction of motion. We neglect weight and normal force perpendicular to the surface of log as they do not affect the motion in x-direction. Now, considering law of motion for the man and log system :

$$\begin{array}{l}{\mathbf{F}}_{\mathrm{Ext.}}=(m+M){\mathbf{a}}_{\mathrm{COM}}\\ \Rightarrow {\mathbf{a}}_{\mathrm{COM}}=0\end{array}$$

Thus, COM is not accelerated. Also as the system was at rest in the beginning, COM of the system at the end should remain where it was in the beginning. Now, considering measurement of COM in horizontal direction (x - direction) as measured from bank for the initial condition, we have :

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

$$\begin{array}{l}\Rightarrow x_{\mathrm{COMi}}=\frac{mL+MX\frac{L}{2}}{m+M}\end{array}$$

Let the log has moved off "x" distance when the person reaches the nearer end. The COM in horizontal direction (x - direction) as measured from the bank for the final condition is :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COMf}}=\frac{mx+MX(\frac{L}{2}+x)}{m+M}\end{array}$$

But,

$$\begin{array}{l}{x}_{\mathrm{COMi}}=x_{\mathrm{COMf}}\end{array}$$

$$\begin{array}{l}\Rightarrow mL+MX\frac{L}{2}=mx+MX(\frac{L}{2}+x)\end{array}$$

$$\begin{array}{l}\Rightarrow x=\frac{mL}{m+M}\end{array}$$