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Standard XII

Physics

COM of Center of Masses

Position of t...

Question

Position of two particles are given by $x_{1} = 2 t$ and $x_{2} = 2 + 3 t$ . The velocity of centre of mass at $t = 2 s$ is $2 m/s$ . Velocity of centre of mass at $t = 4 s$ will be

2 m/s

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4 m/s

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1 m/s

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zero

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Solution

The correct option is A $2 m/s$
Since acceleration of the given particles are $a_{1} = \frac{d^{2} x_{1}}{d t^{2}} = 0, a_{2} = \frac{d^{2} x_{2}}{d t^{2}} = 0$
$∴$ Velocity of centre of mass remains constant
$∴$ (a).

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COM of Center of Masses

Standard XII Physics

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Position of two particles are given by x1=2t and x2=2+3t. The velocity of centre of mass at t=2 s is 2 m/s. Velocity of centre of mass at t=4 s will be

The correct option is A 2 m/sSince acceleration of the given particles are a1=d2x1dt2=0 , a2=d2x2dt2=0 ∴ Velocity of centre of mass remains constant ∴ (a).

Position of two particles are given by $x_{1} = 2 t$ and $x_{2} = 2 + 3 t$ . The velocity of centre of mass at $t = 2 s$ is $2 m/s$ . Velocity of centre of mass at $t = 4 s$ will be

The correct option is A $2 m/s$
Since acceleration of the given particles are $a_{1} = \frac{d^{2} x_{1}}{d t^{2}} = 0, a_{2} = \frac{d^{2} x_{2}}{d t^{2}} = 0$
$∴$ Velocity of centre of mass remains constant
$∴$ (a).