Two equal lumps of putty are suspended side by side from two long string 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 other one. the vertical distance risen by the centre of gravity of the combination is?

Question

Answer

Situation Overview
Two equal lumps of putty are hanging from strings, touching each other. One lump is pulled aside, raising its center of gravity by a distance $ h $, and then released. Upon colliding with the stationary lump, they stick together, forming a single mass.
Energy Considerations
- When the first putty lump is raised, it gains potential energy given by:
$$ PE = mgh $$
where $ m $ is the mass of the putty, $ g $ is the acceleration due to gravity, and $ h $ is the height gained.
Collision Dynamics
- The collision is inelastic, meaning the two lumps stick together after the collision.
- Before the collision, the velocity $ v $ of the first lump can be derived from its potential energy:
$$ \frac{1}{2} mv^2 = mgh $$
Thus,
$$ v = \sqrt{2gh} $$
Conservation of Momentum
- The total momentum before the collision equals the total momentum after the collision:
Let $ V $ be the velocity of the combined mass after the collision.
$$ mv = (2m)V $$
Simplifying gives:
$$ V = \frac{v}{2} = \frac{\sqrt{2gh}}{2} = \frac{\sqrt{gh}}{\sqrt{2}} $$
Height Calculation After Collision
- The new combined mass of the putty rises to a height $ h' $ after the collision. Using the potential energy equation:
$$ \frac{1}{2}(2m)g h' = \frac{1}{2}mv^2 $$
Simplifying leads to:
$$ 2gh' = 2gh $$
Thus,
$$ h' = \frac{h}{2} $$
Conclusion
The vertical distance risen by the center of gravity of the combined putty lumps after the collision is $ \frac{h}{2} $.

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