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A particle of mass 10 kg is moving with ...

A particle of mass `10 kg` is moving with a uniform speed of `6 m//sec` in `x-y` plane along the line `3y = 4x + 10`, what is the magnitude of its angular momentum about the origin in `kg - m^(2)//s` ?

Text Solution

Verified by Experts

The correct Answer is:
`120`
`3y = 4x + 10`
`|vec p| = mv = 10 xx 6 = 60 kg m//s`
The `bot` distance form origin on line
=`(-0' + 0 +10)/(sqrt((3)^(2) + (4)^(2))) = (10)/(56) = 2m`
`|L| =P xx bot dist = (60)(2) = 120 kg -m^(2)//s`.
Alternatively
`tan theta = 4//3`
`theta = 53^@`
`r = (5)/(2) sin 53^@ = 2m`
`L = rP rArr L = rmv`
`L = 2 xx 10 xx 6`
`L = 120 Kg-m//s`.
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