Q

# 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)

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

Views

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)

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

Exams
Articles
Questions