Question

# A particle of mass 2 kg is projected on horizontal ground with an initial velocity u=20 m/s making an angle 60∘ with the vertical. Find out the angular momentum of the particle about the point of projection when it just strikes the ground. [in kg m2/s]

A
2003
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B
1003
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C
zero
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D
4003
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Solution

## The correct option is D 400√3Given, Mass of particle, m=2 kg Initial velocity of particle, u=20 m/s Assume particle hits the ground after time t at point P. Vertical component of initial velocity =uy uy=usinα=usin30∘=20×12=10 m/s As we know, vertical component of particle's speed, at the time of projection and when it hits the ground will be same. Hence, vy=10 m/s As we can see, at point P, horizontal component of velocity passes through the origin i.e r⊥=0, Hence, there is no angular momentum due to horizontal component of particle's velocity) Angular momentum because of vertical component of velocity : L=mvyr⊥ Here r⊥ = Range of projectile = R R=u2sin2αg=202×sin(60∘)10=20√3 m Hence L=(mvy×R)=2×10×20√3 L=400√3 kg m2/s

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