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Help me please, - Laws of motion - JEE Main

A wedge of mass M=4m lies on a frictionless plane . A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge . The maximum height climbed by the particle on the wedge is given by :

  • Option 1)

    \frac{v^{2}}{g}

  • Option 2)

    \frac{2v^{2}}{7g}

  • Option 3)

    \frac{2v^{2}}{5g}

  • Option 4)

    \frac{v^{2}}{2g}

 
Answers (1)
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Applying  Conservation of linear momentum

m \:V_{o}=(m+nm)V

\Rightarrow r=\frac{V_{0}}{5}

Applying conservation of mechanical energy

\frac{1}{2}mV_{0}^2=\frac{1}{2}(m+nm)V^{2}+mgH

gH=\frac{1}{2}\left [ V_{0}^{2}-5\frac{V_{0}^{2}}{25} \right ]

\Rightarrow H= \frac{2}{5}\frac{V_{0}^{2}}{g}


Option 1)

\frac{v^{2}}{g}

Option 2)

\frac{2v^{2}}{7g}

Option 3)

\frac{2v^{2}}{5g}

Option 4)

\frac{v^{2}}{2g}

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