91Ó°ÊÓ

i) Mass of the one particle is 2.0 kg

ii) Mass of the other particle is 4.0 kg

iii) Velocity of the one particle is,$\stackrel{→}{v_1}=-4\stackrel{Áåœ}{i}-5\stackrel{Áåœ}{j}$

iv) Velocity of the other particle is,$\stackrel{→}{v_2}=6\stackrel{Áåœ}{i}-2\stackrel{Áåœ}{j}$

Using the formula for the velocity vector of the center of mass, find the velocity vector of the center of mass of particles P and Q. From this, find its magnitude and direction.

Formula is as follow:

${\stackrel{→}{v}}_{com}=\frac{\left({\stackrel{→}{v}}_1{m}_1+{\stackrel{→}{v}}_2{m}_2\right)}{\left(m_1+m_2\right)}$

Here, m₁, m₂ are masses, v₁, v₂ are velocities of masses m₁, m₂ and v_cm is velocity of centre of mass.

Use following formula to calculate velocity unit vector notation,

$$\begin{gathered} {\stackrel{→}{v}}_{com}=\frac{\left({\stackrel{→}{v}}_1{m}_1+{\stackrel{→}{v}}_2{m}_2\right)}{\left(m_1+m_2\right)} \\ {v}_{com}=\frac{\left(-4\stackrel{Áåœ}{i}-5\stackrel{Áåœ}{j}\right)2+\left(6\stackrel{Áåœ}{i}-2\stackrel{Áåœ}{j}\right)4}{4+2} \\ {v}_{com}=2.66666\stackrel{Áåœ}{i}-3\stackrel{Áåœ}{j}≈2.67\stackrel{Áåœ}{i}-3\stackrel{Áåœ}{j} \end{gathered}$$

Hence, velocity in unit vector notation is $2.67\stackrel{Áåœ}{i}-3\stackrel{Áåœ}{j}$.

Magnitude of velocity is as follows:

$$\begin{gathered} v_{com}=\sqrt{2.{66666}^2+{\left(-3\right)}^2} \\ {v}_{com}=4.01\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, magnitude of velocity is $4.01\frac{m}{s}$.

Angle is,

$$\begin{gathered} \tan \mathrm{θ}=\frac{v_y}{v_x} \\ \tan \mathrm{θ}=-\frac{3}{2.66666} \\ \mathrm{θ}=-48.4° \end{gathered}$$

Hence,angle is$-48.4°$.

From the masses and velocities of objects, the magnitude and direction of the velocity of their center of mass can be found.