Question bank/JEE Main/Physics

An object A is dropped from rest from the top of a 30 m high
building and at the same moment another object B is projected vertically
upwards with an initial speed of 15 m/s from the base of the building.
Mass of the object A is 2 kg while mass of the object B is 4kg. The
maximum height above the ground level attained by the centre of mass of
the A and B system is (take$g = 10m/s^{2}$)

A15 mB25 mC30 mD35 m

Show answer and solution▾

Using
$\frac{m_{1}{\overset{\rightarrow}{x}}_{1} + m_{2}{\overset{\rightarrow}{x}}_{2}}{m_{1} + m_{2}} = X_{cm}$
By putting values $X_{cm} = 10\ m$from base.
[IMAGE] Initial velocity of
COM$\Rightarrow \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = 10m/supward$
Also, acceleration of COM is g thus using
$v^{2} = u^{2} + 2$as for COM
$0 = 10^{2} + 2 \times ( - 10) \times h_{\max}$
$h_{\max} = 5m$from the initial position of COM
Hence total height from base$= 10 + 5m = 15m$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx