QJEEM20Phd4f6: JEE Main 2020 Physics - Class 11 JEEM20Phd4f6 Solution: A particle of mass m is dropped from a height h above the…

Particles will collide after time t = ${h \over {\sqrt {2gh} }}$ = $\sqrt {{h \over {2g}}}$

Velocity of (A) just before collision,
V_A = 0 + gt = $g\sqrt {{h \over {2g}}}$ = $\sqrt {{{gh} \over 2}}$

Velocity of (B) just before collision,
V_B = ${\sqrt {2gh} }$ - $g\sqrt {{h \over {2g}}}$

= ${\sqrt {2gh} }$ - $\sqrt {{{gh} \over 2}}$

= $\sqrt {gh} \left[ {\sqrt 2 - {1 \over {\sqrt 2 }}} \right]$

As 'mg' is non-impulsive so conserving linear momentum just before and just after collision,

P_i = P_f

$\Rightarrow$ m $\sqrt {gh} \left[ {\sqrt 2 - {1 \over {\sqrt 2 }}} \right]$ - m $\sqrt {{{gh} \over 2}}$ = $2m{V_f}$

$\Rightarrow$ V_f = 0

height from ground where collision
takes place = h₁ = h - ${1 \over 2}g{t^2}$ = h - ${1 \over 2}g.{h \over {2g}}$ = ${{3h} \over 4}$

Time taken by combined mass to reach the
ground = $\sqrt {{{2 \times {{3h} \over 4}} \over {2g}}}$ = $\sqrt {{{3h} \over {2g}}}$