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On a frictionless surface, a block of mass M moving at speed $\upsilon$ collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $\frac{\upsilon } {3}$ . The second block's speed after the collision is :

Option 1)
$\frac{3} {4}\upsilon$

Option 2)
$\frac{3} {{\sqrt 2 }}\upsilon$

Option 3)
$\frac{{\sqrt 3 }} {2}\upsilon$

Option 4)
$\frac{{2\sqrt 2 }} {3}\upsilon$

Answers (1)

As we discussed in

Conservation of Energy and Conservation of Mechanical Energy -

Accounting all forms of energy within isolated system

i.e. total energy remains constant.

-

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$frac{1}{2}m_{1}u_{1}^{2}+frac{1}{2}m_{2}u_{2}^{2}= frac{1}{2}m_{1}v_{1}^{2}+frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial : and: final : velocity: of : the: mass m_{1}$

$u_{2},v_{2}:initial : and: final : velocity: of : the: mass m_{2}$

According to conversation of energy

$\frac{1}{2}mv^{2}+0 =\frac{1}{2}m\left ( \frac{v}{3} \right )^{2}+\frac{1}{2}mv^{2}$

$v^{2}=\frac{v^{2}}{9}+v^{'^{2}}$

$= v^{'^{2}}=v^{2}-\frac{v^{2}}{9}$

$=\frac{8}{9}v^{2}$

$= v^{'^{2}}=\frac{8}{9}v^{2}$

$v^{'}=\frac{2\sqrt{2}}{3}v$

Option 1)

$\frac{3} {4}\upsilon$

Incorrect option

Option 2)

$\frac{3} {{\sqrt 2 }}\upsilon$

Incorrect option

Option 3)

$\frac{{\sqrt 3 }} {2}\upsilon$

Incorrect option

Option 4)

$\frac{{2\sqrt 2 }} {3}\upsilon$

Correct option

Posted by

Vakul

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