Question

$\huge{\underline{\underline{\sf{Question-}}}}$

Prove that if the total external force acting on a system of particles is zero, then velocity of the centre of mass remains constant.

Thank you!​

Answer 1

AnswEr:

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We know that,

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$\sf{\vec{P} = m \vec{v}}$

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Where,

• P is linear momentum.

• m is mass.

• v is velocity.

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Also, using second law of motion,

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$\sf{\vec{F} = \dfrac{d \vec{P}}{dt}}$

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That is Rate of change of linear momentum of a particle is equal to the net force acting on the object.

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$\sf{M \vec{v_{CM}} = m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}} + \cdots + m_{n} \vec{v_{n}}}$

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Using above equation, we can write -

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$\sf{M \vec{v_{CM}} = \large\displaystyle \sum\limits_{i=1}^{n} m_{1} \vec{v_{1}}}$

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And, $\sf{\displaystyle \sum\limits_{i=1}^{n} \vec{P_{1}} = \displaystyle \sum\limits_{i=1}^{n} m_{1} \vec{v_{1}}}$

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Here, the left hand side is the summation of linear momentum of n particles of the system, which is equal to the product of the total mass of the system and velocity of the center of mass of the system.

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$\longrightarrow$ $\sf{\vec{P} = M \vec{v_{CM}} ----------- (i)}$

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†Differentiating with respect to time,

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$\sf{\dfrac{d \vec{P}}{dt} = M \dfrac{d \vec{v}_{CM}}{dt} = M \vec{a}_{CM} = \vec{F}_{external}}$

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If external force is zero, linear momentum of the system is conserved and the centre of mass will move with constant velocity.

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$\sf{\dfrac{d \vec{P}}{dt} = 0\:and\: \vec{P} =\:constant}$

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$\therefore$ If total external force acting on a system of particles is zero, then velocity of the centre of mass remains constant.

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Hence proved!

Answer 2

To prove:- if the total external force acting on a system of particles is zero, then velocity of the centre of mass remains constant.

Proof :-

We know that Linear momentum of particles is written as:-

$\implies \: p = mv$

where P = momentum , m = mass and v = Velocity

Newton's second law of single particles is written as,

$\implies \: F = \frac{dp}{dt}$

where F stands for the force on the particles. ( Yaad aaya :')?)

Now let's assume n number of particles with masses m1 , m2, m3,... mn and also their Velocity be v1, v2 , v3 ....Vn respectively .

Since the particles may be interacting and have some external forces so the linear momentum of first particle becomes m1v1 for second particle it becomes m2v2 , for third particles it becomes m3v3 and so on.

Now, Linear momentum of all the individual particles will be ,

$\implies \: P = p1 \: + \: p2 + p3 + .... + pn \\ = m1v1 + m2v2 + .... + mnvn$

Remember P = MV [ where MV = m1v1 + m2v2 + ...mn]

We also know that the total momentum of the particles is equal to the product of total mass of the particles and the velocity of it's centre of mass.

So by differentiating it with the respect to time ,

$\ \frac{dp}{dt} = M \frac{dv}{dt} =MA$

Since dv/dt is the acceleration of the centre of mass so,

$\implies\frac{dp}{d} = Fext$

The equation that we got above is nothing but the Newton’s second law to a system of particles .

And now suppose if the external force acting is zero then ,

$\implies \frac{dp}{dt} = 0$

or simply P = constant .

Hence proved!! ( kitta easy peasy tha na :')?)