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A particle of mass m is projected at 45 degrees to the horizontal with initial speed V0 from point P at time t = 0. What is the magnitude of the angular momentum of the particle about the point P at time t = V0/g?
- (1 / (2*sqrt(2))) * m * V0³ / g
- (1 / (2*sqrt(2))) * m * V0² / g
- (1/2) * m * V0³ / g
- (1/2) * m * V0² / g
Correct answer: (1 / (2*sqrt(2))) * m * V0³ / g
Solution
At t = V0/g: x = (V0/sqrt(2)) * (V0/g) = V0²/(g*sqrt(2)). y = (V0/sqrt(2))*(V0/g) - (1/2)*g*(V0/g)² = V0²/(g*sqrt(2)) - V0²/(2g). vₓ = V0/sqrt(2), v_y = V0/sqrt(2) - g*(V0/g) = V0*(1/sqrt(2) - 1). Angular momentum L = m * |r x v| = m * |x*v_y - y*vₓ|. Computing gives L = m*V0³/(2*sqrt(2)*g).
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