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Question

# A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time t<(v0 sin θ/g), the angular momentum of the particle is

A
mgv0t2 cos θ ^j
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B
mgv0t cos θ ^k
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C
12 mgv0t2 cos θ ^k
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D
12 mgv0t2 cos θ ^i
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Solution

## The correct option is C 12 mgv0t2 cos θ ^kThe position vector of the particle from the origin at any time t can be calculated by using →r=→ut+12→at2 ⇒→r=v0 cos θ t →i+(v0 sin θ t−12 g t2)^j ∴ Velocity vector at any time can be calculated by using →v=→u+→a t, ⇒→v=v0 cos θ^i+(v0 sinθ−gt)^j The angular momentum of the particle about the origin is →L=→r×m→v →L=m(→r×→v) =m[v0 cosθt^i+(v0 sinθt−12gt2)^j)×(v0 cosθ^i+(v0 sin θ−gt)^j)] =m[v20 cosθ sinθt−v0gt2 cosθ)^k+(v20 sin θ cosθt−12gt2v0cosθ)(−^k)] =m[v20 sinθ cosθt^k−v0gt2 cosθ^k−(v20 sin θ cosθt^k+12v0gt2cosθ^k] =m[−12v0gt2 cosθ^k]=12mgv0t2 cosθ^k

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