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Standard XII

Physics

Center of Mass as an Average Point

Two charged p...

Question

Two charged particles of masses $m$ and $2 m$ are placed a distance $d$ apart on a smooth horizontal table. Because of their mutual attraction, they move towards each other and collide. Where will the collision occur with respect to the initial positions?

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Solution

Since the bodies move under mutual attraction of one-other and there is no external force on the system, the CM will not move, And hence the masses will collide at the Center of Mass.

Now, distance of CM from mass $m$ is $x = \frac{2 m}{m + 2 m} d$

$⟹ x = \frac{2 d}{3}$ from $m$

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Two charged particles of masses m and 2m are placed a distance d apart on a smooth horizontal table. Because of their mutual attraction, they move towards each other and collide. Where will the collision occur with respect to the initial positions?

Since the bodies move under mutual attraction of one-other and there is no external force on the system, the CM will not move, And hence the masses will collide at the Center of Mass.Now, distance of CM from mass m is x=2mm+2md⟹x=2d3 from m

Two charged particles of masses $m$ and $2 m$ are placed a distance $d$ apart on a smooth horizontal table. Because of their mutual attraction, they move towards each other and collide. Where will the collision occur with respect to the initial positions?

Since the bodies move under mutual attraction of one-other and there is no external force on the system, the CM will not move, And hence the masses will collide at the Center of Mass.

Now, distance of CM from mass $m$ is $x = \frac{2 m}{m + 2 m} d$

$⟹ x = \frac{2 d}{3}$ from $m$