Two particles A and B of masses 1kg and 2kg respectively are projected in the directions shown in figure with speeds `u_A=200m//s` and `u_B=50m//s`. Initially they were `90m` apart. They collide in mid air and stick with each other. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant.
`(g=10m//s^2)`

Answer

A

Explanation

Using `m_Ar_A=m_Br_B`
or `(1)(r_A)=(2)(r_B)`
or `r_A=2r_B`
and `r_A+r_B=90m`
Solving these two equations, we get
`r_A=60m` and `r_B=30m`
i.e. COM is at height 60 m from the ground at time `t=0`.
Further, `a_(COM)=(m_Aa_A+m_ga_B)/(m_A+m_B)`
`=g=10m//s^2` (downwards)
as `a_A=a_g=g` (downwards)
`u_(COM)=(m_Au_A+m_Bu_B)/(m_A+m_B)`
`=((1)(200)-(2)(50))/(1+2)=(100)/(3)m//s` (upwards)
Let, h be the height attained by COM beyond 60m. Using,
`v_(COM)^2=u_(COM)^2+2a_(COM^h)`
or `0=(100/3)^2-(2)(10)h`
or `h=(100)^2/180=55.55m`
Therefore, maximum height attained by the centre of mass is
`H=60+55.55=115.55m`