Two particles of mass `m_(1)` and `m_(2)` are projected from the top of a tower. The particle `m_(1)` is projected vertically downward with speed u and `m_(2)` is projected horizontally with same speed. Find acceleration of CM of system of particles by neglecting the effect of air resistance.

Two particles of mass `m_(1)` and `m_(2)` are projected from the top o

1.	Two particles of mass `m_(1)` and `m_(2)` are projected from the top of a tower. The particle `m_(1)` is projected vertically downward with speed u and `m_(2)` is projected horizontally with same speed. Find acceleration of CM of system of particles by neglecting the effect of air resistance.
Answer» As effect of air is neglected, therefore the only force acting on the particles is the gravitational force is downward direction. Let the point of projection is taken as origin and downward direction as negative Y-axis, then acceleration of 1st point mass `a_(1) = -gj` acceleration of 2nd point mass `a_(2)` = `-gj` `a_(CM)` = `(m_(1)a_(1) `+ `m_(2)a_(2))/(m_(1) + m_(2)` = `(m_(1)(-gj) + m_(2)(-gj))/(m_(1) + m_(2)` `rArr a_(CM) = -gj` i.e., acceleration of CM is equal to acceleration due to gravity and is in downward direction. Note. If large number of particles are projected under the effect of gravity only indifferent direction, then acceleration of CM is equal to the acceleration due to gravity irrespective of direction of projection of particles.

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Two particles of mass `m_(1)` and `m_(2)` are projected from the top of a tower. The particle `m_(1)` is projected vertically downward with speed u and `m_(2)` is projected horizontally with same speed. Find acceleration of CM of system of particles by neglecting the effect of air resistance.

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