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Question

# A particle of mass 200 g is projected from origin O with speed 10 m/s at an angle 60∘ with positive x− axis. Find the magnitude of angular momentum of particle after 1 sec about the origin. (Take g=10 m/s2)

A
25 kg-m2/sec
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B
0 kg-m2/sec
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C
250 kg-m2/sec
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D
125 kg-m2/sec
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Solution

## The correct option is D 125 kg-m2/secGiven, mass of the particle m=200 g Speed of the particle u=10 m/s θ=60∘,t=1 sec Let the angular momentum of the particle after t=1 sec is →L Angular momentum is given by →L=m(→r×→v) Now, →r(t)=x^i+y^j=(ucosθ)t^i+(usinθ×t−12gt2)^j ……(1) where, x and y is the distance travelled by the particle after time (t) in x and y direction respectively. →r(t)=(10cos60∘)×1^i+(10sin60∘×1−12×10×12)^j r(t)=5^i+(5√3−5)^j ……(2) Now, velocity of particle →v(t)=vx^i+vy^j=(ucosθ)^i+(usinθ−gt)^j v(t)=5^i+(5√3−10)^j ……(3) ∴→L=m(→r×→v)=0.2∣∣ ∣ ∣∣^i^j^k5(5√3−5)05(5√3−10)0∣∣ ∣ ∣∣ ⇒→L=0.2[0×^i−^j×0+(−25)^k] →L=−25^k kg-m2/sec So, magnitude of angular momentum |→L|=25 kg-m2/sec

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