Question

A small particle of mass m is projected at an angle ¢ with the x-axis with an initial velocity V₀ in the xy place as shown in the figure . At a time t <V₀sin¢/g , the angular momentum of the particle is ?
[AIEEE :- 2010]​

Answer 1

• Explanation :- Angular momentum of the particle is L = m(r×v)
• => where , r = [(V₀cosθt )i + (V₀sinθt - 1/2gt²)j]
• where , V₀ =(V₀cos¢)i + (V₀sin¢-gt)j

now , we have

• L = m [(V₀cos¢t)i + (v₀sin¢t -1/2gt²)j]×[V₀cos¢i +(V₀sin¢ - gt)j]
• we know i× j = k
• ∴ L = mv₀cosθt(-1/2gt)k
• -1/2mgV₀t²cosθkAnswer

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Answer 2

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