Two equal lumps of putty are suspended side by side from two long strings so that they are just touching. One is drawn aside so that it's centre of gravity rises a vertical distance h. It is released and then collides inelastically with the other one. The vertical distance risen by the centre of gravity of the combination is??



Please ans this fast....

Dear student,

Let the masses be m
the potential energy of 1st body due to height 'h' of the Centre of Mass = mgh
Just Before Colliding,
Total potential energy will be converted into kinetic energy under the conservation of Mechanical Energy
​​​​​​Let the velocity of the mass is v, then

$$\begin{gathered} P{E}_{org}+K{E}_{org} = P{E}_{new} + K{E}_{new} \\ K{E}_{new}+0 =0+P{E}_{org} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}m{v}^2=mgh \\ \Rightarrow v^2 = 2gh \end{gathered}$$​​​​
Since the Collision is inelastic, the two bodies will stick the new body generated will have mass '2m'
 

Let the velocity after the collision is v', then

By conservation of linear momentum

$$\begin{gathered} mv + 0 = 2m{v}^{\text{'}} \\ \Rightarrow v^{\text{'}}=\frac{v}{2} \end{gathered}$$

The kinetic energy of the combination
$$\begin{gathered} = \frac{1}{2}\left(2m\right){\left[v^{\text{'}}\right]}^2 \\ = \frac{m{v}^2}{4} \\ =\frac{m\left(2gh\right)}{4} \\ = \frac{mgh}{2} \end{gathered}$$


Let the vertical distance risen by the new Centre of Mass  is h'
Here, Again Using Conservation of Mechanical Energy
$$\begin{gathered} P{E}_{org}+K{E}_{org} = P{E}_{new} + K{E}_{new} \\ P{E}_{new}+0 =0+K{E}_{org} \end{gathered}$$
$$\begin{gathered} 2mg{h}^{\text{'}}== \frac{1}{2}\left(2m\right){\left[v^{\text{'}}\right]}^2=\frac{mgh}{2} \\ \Rightarrow h^{\text{'}}=\frac{h}{4} \end{gathered}$$.
​​​​​The vertical distance of Centre Of Mass of combination risen from the ground is h/4

Regards