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Question

A particle of mass m=1 kg is thrown from a height of 20 m from the ground with a horizontal speed 10 m/s. Find the angular momentum [in kg m2/s] of the particle about the point of projection when it hits the ground. (Take g=10 m/s2)


A
Zero
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B
400(^k)
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C
200(^k)
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D
600(^k)
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Solution

The correct option is C 200(^k)
Initially, at time t=0
Horizontal component of speed ux=u=10 m/s
Vertical component of speed uy=0
At time t, when it hits the ground:
(Assume upward motion as positive)
Horizontal speed vx=ux (no horizontal acceleration)
From eq. v2=u2+2as,
v2y=u2y+2ays=02gs
v2y=2×(10)×(20)
or vy=20 m/s
So velocity vector when particle hits the ground,
v=10^i20^j=vx^i+vy^j
Time of flight t can be calculated from,
vy=uy+ayt20=0+10×t
i.e t=2 s
So, distance travelled by the particle:
In horizontal direction, Sx=uxt=20 m
In vertical direction Sy=20 m
Hence, particle will hit the ground at A(20,0).


Position vector from point of projection,
r=(200)^i+(020)^j=20^i20^j
As we know, angular momentum
L=m(r×v)
L=1[(20^i20^j)×(10^i20^j)]
=400(^i×^j)200(^j×^i)
=400^k200(^k)
L=200^k

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