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question is in (attached image ) plz solve my question as soon as possible

Sarvesh , 5 Years ago
Grade 12
anser 2 Answers
Khimraj

Last Activity: 5 Years ago

velocity of of sticked object at lowest point = \sqrt{2gh}/2
so max height = u2/2g = h/2
Hope it clears...................................

Rajdeep

Last Activity: 5 Years ago

HELLO THERE!
I’m trying to upload an image below, hope it reaches you:
 
 
So as in the above diagram (which you will see if it reaches you), we see that the radius of the sphere is h/2, since the height is h.
 
Now, we know, that if the radius is r, then  velocity of any particle at the bottom is 
\sqrt{2gr}
 
Here,
r = \frac{h}{2},
 
so we get velocity of the moving particle at the bottom as:
v = \sqrt{2g\frac{h}{2}}
 
So with this velocity, the moving particleof mass m strikes the particle at rest.
Applying Conservation of Linear Momemtum, we can calculate the equal final velocity, with which they move on!
 
Initial momentum = Final momentum (let their final velocity be v)
m\times \sqrt{2g\frac{h}{2}} = 2mv \\\\\implies v = \sqrt{2g\frac{h}{2}}
 
So, both the particles together move with the above velocity. With this,we can calculate the total height h’ to which the system of blocks rise:
 
h^{\prime} = \frac{v^{2}}{2g} \\\\= \frac{2g\frac{h}{2}}{2g} \\\\=\frac{h}{2}
 
HOPE THAT HELPS!

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