Question:
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)
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)
(Take g = 10 m/s2)
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Find the speed at B by energy conservation (\(h=10\) m), then use \(L=mvr\) with \(r=OB=a+h=20\) m, since the velocity at the lowest point B is perpendicular to OB.
Updated On: Nov 5, 2026
- 6 kg-m2/s
- 8 kg-m2/s
- 2 kg-m2/s
- 3 kg-m2/s
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The Correct Option is A
Solution and Explanation
Mass \(m=20\) g \(=0.02\) kg; figure gives \(h=10\) m and \(a=10\) m (O is 10 m above the level of A).
Energy conservation A to B: \(\frac12mv_B^2=\frac12mv_A^2+mgh\Rightarrow v_B^2=25+2\times10\times10=225\Rightarrow v_B=15\) m/s.
At B the velocity is horizontal and OB is vertical with \(OB=a+h=20\) m, so \(L=mv_B\,(OB)=0.02\times15\times20=6\) kg-m2/s.
Answer: (1) 6 kg-m2/s
Energy conservation A to B: \(\frac12mv_B^2=\frac12mv_A^2+mgh\Rightarrow v_B^2=25+2\times10\times10=225\Rightarrow v_B=15\) m/s.
At B the velocity is horizontal and OB is vertical with \(OB=a+h=20\) m, so \(L=mv_B\,(OB)=0.02\times15\times20=6\) kg-m2/s.
Answer: (1) 6 kg-m2/s
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