Question

Two particles A and B of mass 1 kg and 2 kg respectively are projected in the direction shown in figure with speeds `u_(A) = 200 ms^(-1)` and `u_(B) = 50 ms^(-1)`. Initially they were 90 m apart. Find the maximum height attained by the center of mass of the particles. Assume acceleration due to gravity to be constant. (Take, `g=10 ms^(-2)`.

Answer 1

Using, `m_(A)r_(A) = m_(B)r_(B)` or `1(r_(A)) = 2(r_(B))`
or `r_(A) = 2r_(B)`..................(i)
and `r_(A) + r_(B)` = 90 m ..........................(ii)
Solving thest two equations, we get
`r_(A)`=60 m and `r_(B)` =30 m
i.e., CM is at height 60 m from the ground at time t=0.
Further, `a_(CM)` = `(m_(A)a_(A) + m_(B)a_(B))/(m_(A) + m_(B))`= g = `10 ms^(-1)` (downwards)
as `a_(A)` = `a_(B`) = g (downwards)
`u_(CM)` = `(m_(A)u_(A) + m_(B)u_(B))/(m_(A) + m_(B)`
`(1(200) - 2(50))/(1+2) = 100/3 ms^(-1)` (upwards)
Let h be the height attained by CM beyond 60m. Using
`v_(CM)^(2) = v_(CM)^(2) + 2a_(CM^(h)`
or `0=(100/3)^(2) -(2)(10(h)`
or `h=(100)^(2)/180= 55.55m`
Therefore, maximum height attained by the center of mass is H=60 + 55.55 = 115.55m