Question

A body of mass $ m $ is dropped and another body of mass $ M $ is projected vertically up with speed $ u $ simultaneously from the top of a tower of height $ H $ . If the body reaches the highest point before the dropped body reaches the ground, then the maximum height raised by the centre of mass of the system from the ground is
A) $ H + \dfrac{{{u^2}}}{{2g}} $
B) $ \dfrac{{{u^2}}}{{2g}} $
C) $ H + \dfrac{1}{{2g}}{\left( {\dfrac{{Mu}}{{m + M}}} \right)^2} $
D) $ H + \dfrac{1}{{2g}}{\left( {\dfrac{{mu}}{{m + M}}} \right)^2} $

Answer 1

Hint : We will use the formula of the velocity of the centre of mass for a system of particles. Since the body of mass $ M $ reaches the highest point before the body reaches the ground, then the centre of mass will reach its highest point which will have a height determined by the third law of kinematics.

Formula used: In this question, we will use the following formula
 $ {v_{COM}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{v_i}} }}{{\sum\limits_{}^{} {{m_i}} }} $ where $ {v_{COM}} $ is the velocity of the centre of mass of the system, $ {m_i} $ is the mass and $ {v_i} $ is the velocity of the $ {i^{{\text{th}}}} $ object.
 $ {v^2} = {u^2} - 2gh $ where $ v $ is the final velocity of the centre of mass, $ u $ is the initial velocity, $ g $ is the gravitational acceleration, and $ h $ is its height.

Complete step by step answer
We’ve been told that a body of mass $ m $ is dropped and another body of mass $ M $ is projected vertically up simultaneously from the top of the tower. As these masses drop and rise, the centre of mass will change accordingly and it will shift with a certain velocity.
At the starting time, the body of mass $ m $ is stationary while the body of mass $ M $ has speed $ u $ . We can calculate the velocity of the centre of mass at the starting time using
 $ {v_{COM}} = \dfrac{{\sum\limits_{i = 1}^n {{m_i}{v_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} $
So, for our case
 $ {v_{COM}} = \dfrac{{m(0) + M(u)}}{{m + M}} $
 $ \Rightarrow {v_{COM}} = \dfrac{{Mu}}{{m + M}} $
The velocity of the centre of mass will move with this velocity under a retardation due to gravitational acceleration $ g $ . The maximum height can then be achieved using the equation
 $ {v^2} = {u^2} - 2gh $
At the highest point of the centre of mass, it will have zero velocity so we can say $ v = 0 $ and hence we can write
 $ {u^2} = 2gh $
As $ u = {v_{COM}} $ is the initial velocity of the centre of mass, the maximum height obtained by the centre of mass will be
 $ h = \dfrac{1}{{2g}}{\left( {\dfrac{{Mu}}{{m + M}}} \right)^2} $
This will be the height of the centre of mass above the building of height $ H $ hence its height from above the ground will be
$ h = H + \dfrac{1}{{2g}}{\left( {\dfrac{{Mu}}{{m + M}}} \right)^2} $ which corresponds to option (C).

Note
To solve such questions, we should not treat both the masses individually but rather only work with the centre of mass of the two bodies as it will be more simplified. Also, since we’ve been asked to find the highest point above the ground and not form the top of the building, we must add the height of the building to the maximum height of the centre of mass.