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Grade 12Mechanics

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

Question image for question is in (attached image ) plz solve my que

Sarvesh

7 Years agoGrade 12

2 Answers

Khimraj

Approved Tutor Answer7 Years ago

velocity of of sticked object at lowest point = $\sqrt{2gh}/2$

so max height = u²/2g = h/2

Hope it clears...................................

Rajdeep

Approved Tutor Answer7 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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