1.

A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection, when the particle is at its maximum height h, is

Answer»

zero
`(mv^(3))/(4sqrt2g)`
`(mv^(3))/(sqrt2g)`
`msqrt(2gh^(3))`

Solution :About a given point, angular momentum = moment of linear momentum
Height of a projectile= h

`thereforeh=(V^(2)sin^(2)theta)/(2g)=(v^(2)xxsin^(2)45^(@))/(2g)=(v^(2))/(4g)` ….(i)
At maximum height, the velocity of projectile
`vcostheta=vcos45^(@)=v/(sqrt2)`
`therefore` Momentum at highest point `=(mv)/(sqrt2)`
`therefore` Moment of momentum about point O`=(mv)/(sqrt2)xxh`
`therefore` Angular momentum about O =`(mvh)/(sqrt2)`
`thereforeL=(mvh)/(sqrt2)` ...(II)
Put h from (i) in (ii), we get,
`L=(mv)/(sqrt2)XX((v^(2))/(4g))=(mv^(3))/(4sqrt2g)` ...(iii)
From (i), eliminate v in (iii)
Angular momentum `=(mv^(3))/(4sqrt2g)=(mxx(4gh)^(3//2))/(4sqrt2g)`
or `L=msqrt(2gh^(3))` ...(iv)
Options (b, d) REPRESENT the ANSWERS.


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