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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

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B

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C

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D

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Solution

The correct option is B


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

vx = v0 cos 45 = v02

and x = vxt = v02.v0g = v202g.

For vertical motion,

vy = v0 sin 45 gt = v02 v0 = (12)2v0

and y = (v0 sin 45)t12gt2

= v202g - v202g = v202g(21)

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

L = rxp = mrxv

= m(ix + jy)x(ivx + jvy)

= m(kxvy kyvx)

= mk[(v202g)v02(1 2) v202g(2 1)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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