Question 151: 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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