Question

velocity of centre of mass of system of two point masses m and 2m shown in figure is​

Answer 1

Let the centre of mass of the system after collision lies at point C'.

Position of point C w.r.t point A is 3a.

Position of C' from A,      x=m+8m+2mm(a)+8m(3a)+2m(4a)=3a

⟹ Points C and C' coincide.

For translational motion :

As no external force is acting on the system, thus linear momentum is conserved i.e  Pi=Pf

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Answer 2

Answer:

The velocity of centre of mass of the system is $-\frac{v_0}{3}$

Explanation:

• First object has mass, $m_1$ = m

        Its velocity, $v_1$ = $v_0$

       Momentum = $m_1$$v_1$ = m$v_0$

• Second object has mass, $m_2$ = 2m

       Its velocity, $v_2$ = - $v_0$  [as this is moving in opposite direction]

       Momentum of the object = $m_2v_2$ = -2m$v_0$

• As they are coming from opposite direction, so after some time they must collide.

      After collision total mass of the system = $m_1$ + $m_2$ = m + 2m = 3m

      After collision total momentum = $m_1$$v_1$ + $m_2v_2$ = m$v_0$ + (-2m$v_0$)

                                                                                  = m$v_0$ - 2m$v_0$

                                                                                   = -m$v_0$

     Now the velocity of centre of mass, V =  $\frac{m_1v_1+m_2v_2}{m_1+m_2}$

     Putting all the values, V = - $\frac{mv_0}{3m}$ =  $-\frac{v_0}{3}$

∴ The velocity of centre of mass of the system is $-\frac{v_0}{3}$

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