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# A particle of mass 1 kg is projected with a velocity 20√2 m/s from the origin of an x−y co- ordinate axis system at an angle 45∘ with x axis (horizontal). The angular momentum [in kg m2/s] of the ball about the point of projection after 1 second of projection is [Take g=10 m/s2 and y - axis is taken as vertical]

A
200^k
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B
100^k
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C
300^j
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D
350^j
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Solution

## The correct option is B −100^kGiven, Mass of particle, m=1 kg Projected velocity u=20√2 m/s Projected angle with horizontal θ=45∘ As we know, Time of flight T=2usinθg=2×20√2×sin45∘10 =2×20√2×1√210=4 s From time of flight, we can say particle will reach maximum height at time t=2 s, and at time t=1 s particle will be going upward. At time t=1 s, co-ordinates of particle: x=ucosθ×t=20√2×cos45∘×1=20 m y=usinθt−12gt2=20√2×sin45∘×1−12×10×12 =15 m Hence, →r=x^i+y^j=20^i+15^j Velocity components at t=1 s. →vy=uy+ayt=usinθ−g×t =20√2×1√2−10×1=10 m/s and →vx=ucosθ=20√2×cos45∘=20 m/s →v=(20^i+10^j) m/s Angular momentum about point O, →L=m(→r×→v) →L=1×[(20^i+15^j)×(20^i+10^j)] →L=200^k+300(−^k)=100(−^k)  Suggest Corrections  0      Similar questions  Explore more