# On a frictionless surface, a block of mass M moving at speed collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle to its initial direction and has a speed . The second block's speed after the collision is : Option 1) Option 2) Option 3) Option 4)

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)

Incorrect option

Option 2)

Incorrect option

Option 3)

Incorrect option

Option 4)

Correct option

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