Two particles Pand Qare released from rest 1.0 m apart. Phas a mass 0.10kg of aand Qa mass of 0.030 kg.Pand Qattract each other with a constant force of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }\mathbf{1}\mathbf{.}{\mathbf{0}}^{\mathbf{-}\mathbf{2}}\mathbf{N}$. No external forces act on the system.(a) What is the speed of the center of mass of Pand Qwhen the separation is 0.50 m?(b) At what distance from P’s original position do the particles collide?

1. Mass of P is,$M_1=0.1\mathrm{kg}$ .
2. Mass of Q is,$M_2=0.3\mathrm{kg}$ .
3. Separation distance between P and Q is 1 m .

Usetheformula for center of mass to findtheposition.Centre of mass is the point, which represent the mean position of a matter of a body.

Formula is as follow:

$X_{cm}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}$

Here, M₁, M₂ are masses, x₁, x₂ are distances of points from centre and X_cm is centre of mass.

As there is no external force acting initially on the stationary particles P and Q, their center of mass does not move.

Hence, speed of the center of mass of P and Q when separation is 0.5 m is$0\frac{\mathrm{m}}{\mathrm{s}}$

Let’s assume that initially particle A is at x = 0 and particle Q is at x = 1.The particles will collide at their center of mass. Then, according to the concept of center of mass,

$$\begin{gathered} X_{com}=\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2} \\ {X}_{com}=\frac{0.1\left(0\right)+(0.3)\left(1\right)}{(0.1+0.3)} \\ {X}_{com}=0.75\mathrm{m} \end{gathered}$$

Hence, distance from P's original position do the particles collide is 0.75 m.

Therefore, the speed of the center of mass from the external force acting on stationary objects can be found. Using the concept of center of mass, the distance at which objects collide can be found.