Question 154: Centre of Mass and Rotational Motion

Question

A particle of mass m is dropped from a height
h above the ground. At the same time another
particle of the same mass is thrown vertically
upwards from the ground with a speed of $\sqrt {2gh} $. If they collide head-on completely
inelastically, the time taken for the combined
mass to reach the ground, in units of $\sqrt {{h \over g}} $ is :

Accepted Answer

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}} $
= $\sqrt {gh} \left[ {\sqrt 2  - {1 \over {\sqrt 2 }}} ight]$
As 'mg' is non-impulsive so conserving linear momentum just before and just after collision,
P
i
= P
f
$ \Rightarrow $ m$\sqrt {gh} \left[ {\sqrt 2  - {1 \over {\sqrt 2 }}} ight]$ - m$\sqrt {{{gh} \over 2}} $ = $2m{V_f}$
$ \Rightarrow $ V
f
= 0
height from ground where collision
takes place = h
1
= 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 	imes {{3h} \over 4}} \over {2g}}} $ = $\sqrt {{{3h} \over {2g}}} $

Answer

$\sqrt {{1 \over 2}} $

Answer

${1 \over 2}$

Answer

$\sqrt {{3 \over 2}} $

Answer

$\sqrt {{3 \over 4}} $

Centre of Mass and Rotational Motion

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