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A particle of mass 1 kg is projected with a velocity 202 m/s from the origin of an xy co- ordinate axis system at an angle 45 with x axis (horizontal). The angular momentum [in kg m2/s] of the ball about the point of projection after 1 second of projection is [Take g=10 m/s2 and y - axis is taken as vertical]

A
200^k
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B
100^k
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C
300^j
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D
350^j
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Solution

The correct option is B 100^k
Given, Mass of particle, m=1 kg
Projected velocity u=202 m/s
Projected angle with horizontal θ=45
As we know, Time of flight T=2usinθg=2×202×sin4510
=2×202×1210=4 s


From time of flight, we can say particle will reach maximum height at time t=2 s, and at time t=1 s particle will be going upward.
At time t=1 s, co-ordinates of particle:
x=ucosθ×t=202×cos45×1=20 m
y=usinθt12gt2=202×sin45×112×10×12
=15 m
Hence, r=x^i+y^j=20^i+15^j

Velocity components at t=1 s.
vy=uy+ayt=usinθg×t
=202×1210×1=10 m/s
and vx=ucosθ=202×cos45=20 m/s
v=(20^i+10^j) m/s

Angular momentum about point O, L=m(r×v)
L=1×[(20^i+15^j)×(20^i+10^j)]
L=200^k+300(^k)=100(^k)

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