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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∘ = v0√2 and x = vxt = v0√2.v0g = v20√2g. For vertical motion, vy = v0 sin 45∘ − gt = v0√2 − v0 = (1−√2)√2v0 and y = (v0 sin 45∘)t−12gt2 = v20√2g - v202g = v202g(√2−1) The angular momentum of the particle at time t about the origin is L = →rx→p = m→rx→v = m(→ix + →jy)x(→ivx + →jvy) = m(→kxvy − →kyvx) = m→k[(v20√2g)v0√2(1 − √2) − v202g(√2 − 1)v0√2] = −→kmv302√2g. Thus, the angular momentum of the particle is mv302√2g in the negative Z - direction, i.e., perpendicular to the plane of motion, going into the plane.

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