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)
V Vakul

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

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