A ball of mass 1 kg is projected with a velocity of
20√2 m/s from the origin of an xy co-ordinate axis
system at an angle 45° with x-axis (horizontal).
The angular momentum [In SI units) of the ball
about the point of projection after 2 s of projection
is (take g = 10 m/s) (y-axis is taken as vertical)
(1) - 400 K (2) 200 i
(4) - 350 j
(3) 300 j
Answers
Answered by
38
Answer: -400 k
initial vertical and horizontal velocities are 20 m/s and 20 m/s
after 2 sec.
vertical and horizontal velocities are 0 m/s and 20 m/s
and y and x displacements are 20 m and 40 m
so about point of projection angular momentum after 2 sec.
= mvxy - mvyx = 1×20×20 - 1×0×40 = 400
Answered by
9
Answer is -400 k
Explanation
initial vertical and horizontal velocities are 20 m/s and 20 m/s
after 2 sec.
vertical and horizontal velocities are 0 m/s and 20 m/s
and y and x displacements are 20 m and 40 m
so about point of projection angular momentum after 2 sec.
= mvxy - mvyx = 1×20×20 - 1×0×40 = 400
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