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Question

A particle of mass m is projected at an angle θ with the Xaxis with an initial velocity v0 in the XY plane as shown in the figure. At a time t<v0sinθg, the angular momentum of the particle about origin is


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
Position vector of particle from origin at time t is,
r=x^i+y^j
where
x=v0cosθ×t
y=v0sinθ×t12gt2
r=(v0cosθ t)^i+(v0sinθ t12gt2)^j
We also know
v=vx^i+vy^j
Where vx and vy are the component of velocity of particle along x and y direction respectively.
vx=v0cosθ
vy=v0sinθgt
v=(v0cosθ)^i+(v0sinθgt)^j
Hence angular momentum of the particle about origin is represented as:
L=r×P=m(r×v)L=m[v0cosθt^i+(v0sinθt12gt2)^j]×[v0cosθ^i+(v0sinθgt)^j]
L=mv20cosθsinθ t ^kmv0cosθg t2 ^kmv20cosθsinθ t ^k(12mgv0cosθ t2)^k
L=12mgv0t2cosθ ^k

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