Applying conservation of energy,

mgh=21mvB2−21mvA2
⇒vB=2gh+vA2
⇒vB=2×10×10+25
⇒vB=15ms−1
Angular momentum about O,
LO=mvB(h+a)
⇒LO=20×10−3×15×20kgm2s−1
⇒LO=6kgm2s−1
JEE Main 2019 — Physics Mechanics
A particle of mass 20g is released with an initial velocity 5ms−1 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=10ms−2)

Held on 12 Jan 2019 · Verified 6 Jul 2026.
3kgm2s−1
2kgm2s−1
6kgm2s−1
8kgm2s−1
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