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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<v0sinθg, the angular momentum of the particle is
1275553_4ac7751c63cf4d80985e5f5850187c67.PNG

A
mgv0t2cosθ^j
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B
mgv0tcosθ^k
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C
12mgv0t2cosθ^k
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D
12mgv0t2cosθ^i
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Solution

The correct option is C 12mgv0t2cosθ^k
As we know that,
angular momentum =m(r×v)
where r=position vector at time t
v=velocity vector at time t

here r=vocosθt^i+(vosinθt12gt2)^j

and, v=vocosθ^i+(vosinθgt)^j

so angular momentum= {vocosθt^i+(vosinθt12gt2)^j}×{vocosθ^i+(vosinθgt)^j}
=12mgvot2cosθ^k


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