JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 8)
A particle of mass 'm' is moving with speed '2v'
and collides with a mass '2m' moving with
speed 'v' in the same direction. After collision,
the first mass is stopped completely while the
second one splits into two particles each of
mass 'm', which move at angle 45° with respect
to the origianl direction.
The speed of each of the moving particle will
be :-
2 $$\sqrt2$$v
v / (2 $$\sqrt2$$ )
v / $$\sqrt2$$
$$\sqrt2$$v
Explanation
Initial momentum · Pi
= 2mv + 2mv = 4 mv
Let v' be the speed of $$l$$ particle_9th_April_Evening_Slot_en_8_1.png)
$$ \therefore 2{{mv} \over {\sqrt 2 }} = 4mv$$
$$ \Rightarrow $$ v' = $${2\sqrt 2 }$$ v
Let v' be the speed of $$l$$ particle
$$ \therefore 2{{mv} \over {\sqrt 2 }} = 4mv$$$$ \Rightarrow $$ v' = $${2\sqrt 2 }$$ v
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