Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x. find the x component of the velocity of the center of mass (vcm)x at that moment. keep in mind that, in general: vx=dx/dt. express your answer in terms of m1, m2, v1x, and v2x.

The center of mass is given with this formula:
[tex]x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}[/tex]
Velocity is:
[tex]v=\frac{dv}{dt}[/tex]
So, for the velocity of the center of mass we have:
[tex]\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\ v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\[/tex]
In our case it is:
[tex]v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}[/tex]

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snehashish65 snehashish65 · Answer 2 · 2021-10-26T11:15:31+02:00

The x component of the velocity of the center of mass is [tex]{\dfrac{m_{1}v_{1} +m_{2}v_{2}}{(m_{1}+m_{2})}[/tex].

Given data:

The x-components of velocities of the blocks are, [tex]v_{1x}[/tex] and [tex]v_{2x}[/tex].

The expression for the center of mass is given as,

[tex]x = \dfrac{ \sum m_{i}x_{i}}{M}[/tex]

Here, M is the total mass of system.

And velocity of center of mass is,

[tex]v=v_{cmx} = \dfrac{dx}{dt}\\\\v=v_{cmx} = \dfrac{\dfrac{(m_{1}x_{1}+m_{2}x_{2})}{(m_{1}+m_{2})}}{dt} \\\\v=v_{cmx} = {\dfrac{(m_{1})}{(m_{1}+m_{2})} \times \dfrac{dx_{1}}{dt} + {\dfrac{(m_{2})}{(m_{1}+m_{2})} \times \dfrac{dx_{2}}{dt}[/tex]

[tex]v=v_{cmx} = {\dfrac{(m_{1})}{(m_{1}+m_{2})} \times v_{1} + {\dfrac{(m_{2})}{(m_{1}+m_{2})} \times v_{2}[/tex]

[tex]v=v_{cmx} = {\dfrac{m_{1}v_{1} +m_{2}v_{2}}{(m_{1}+m_{2})}[/tex]

Thus, the x component of the velocity of the center of mass is [tex]{\dfrac{m_{1}v_{1} +m_{2}v_{2}}{(m_{1}+m_{2})}[/tex].

Learn more about the center of mass here:

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