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Question

A particle is projected at time t =0 from a point P with a speed v0 at an angle of 45 to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t =v0g.


A

mv3022g^j

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B

mv3022g^(j)

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C

mv302g^j

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D

2mv303^(j)

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Solution

The correct option is B

mv3022g^(j)


Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,

vx=v0 cos 45=v02

and x=vxt=v02.v0g=v202g

For vertical motion,

vγ=v0 sin 45gt=v02v0=122v0

and y=(v0 sin 45)t12gt2

y=v202gv202g=v202g(21)

The angular momentum of the particle at time t about the origin is

L=r×p=mr×v

=m(ix+jy)×(ivx+jvy)

=m(kxvykyvx)

=mk[(v202g)v02(12)v202g(21)v02]

=kmv3022g

Thus, the angular momentum of the particle is mv3022g in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.


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