Question

# Two particles are of the same mass m. One particle is dropped from a height of 10 m and the other particle is projected up with a velocity of 10 m/s. If both stick to each other at some height, find the maximum height reached by their COM. [Take g=10 m/s2]10 m5 m1.25 m2.5 m

Solution

## The correct option is C 1.25 mAssume upward motion as positive.  Given, initial velocity of second particle v2=10 m/s upward  Initial velocity of first particle v1=0 m/s downward Let the mass of both particles be m. Acceleration of second particle a2=−g downwards Acceleration of first particle a1=−g downwards Then, velocity of COM VCOM=m1v1+m2v2m1+m2 Vcom=m(0)+m(10)m+m=5 m/s Vcom=5 m/s (upwards) Acceleration of COM =m1a1+m2a2m1+m2 =m(−g)+m(−g)m+m=−2mg2m =−g We can use V22−V21=2aS{HereV1=Vcom} Here, speed of COM at maximum height V2=0 ⇒02−52=2×−10×S ⇒S=1.25 m Hence the maximum height attained by the COM of the two particle system is 1.25 m.

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