Question

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 its 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-
(1) h.
(2) 3h/4
(3) h/2
(4) h/4​

Answer 1

Answer:

Option (4) h/4

Explanation:

Let the masses be m

The potential energy due to height h of the first CoM = mgh

When the first mass will collide with the other mass then at that point h = 0

Total potential energy will be converted into kinetic energy

Let the velocity of the mass is v, then

$\frac{1}{2}mv^2=mgh$

$\implies v^2=2gh$

Since it collides inelastically, therefore the two masses will stick to be a mass of 2m

If the velocity after collision is v' then

By conservation of linear momentum

$mv+0=2mv'$

$\implies v'=v/2$

The kinetic energy of the combination

$=\frac{1}{2}\times 2m(\frac{v}{2})^2$

$=m(\frac{v^2}{4})$

$=m(\frac{2gh}{4})$

$=\frac{mgh}{2}$

Let the vertical distance risen by the other CoM is h'

Then upto height h', total kinetic energy of the combination will be converted into potential energy

Therefore,

$mgh'=\frac{mgh}{2}$

$\implies h'=\frac{h}{4}$

Therefore, the Centre of Mass of the combination will rise to height h/4

Hope this is helpful.