JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 22)
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. when the particle reaches point b, its angular momentum about O will be :
(Take g = 10 m/s2)
_12th_January_Evening_Slot_en_22_1.png)
(Take g = 10 m/s2)
6 kg-m2/s
8 kg-m2/s
2 kg-m2/s
3 kg-m2/s
Explanation
Work Energy Theorem from A to B
Mgh = $${1 \over g}$$ mv$$_B^2$$ $$-$$ $${1 \over g}$$ mv$$_A^2$$
2gh = $$v_B^2 - v_A^2$$
2 $$ \times $$ 10 $$ \times $$ 10 = v$$_B^2$$ $$-$$ 52
vB = 15m/s
Angular momentum about 0
L0 = mvr
= 20 $$ \times $$ 10$$-$$3 $$ \times $$ 20
L0 = 6 kg.m2/s
Mgh = $${1 \over g}$$ mv$$_B^2$$ $$-$$ $${1 \over g}$$ mv$$_A^2$$
2gh = $$v_B^2 - v_A^2$$
2 $$ \times $$ 10 $$ \times $$ 10 = v$$_B^2$$ $$-$$ 52
vB = 15m/s
Angular momentum about 0
L0 = mvr
= 20 $$ \times $$ 10$$-$$3 $$ \times $$ 20
L0 = 6 kg.m2/s
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