1.

A projectile of mass m is thrown with a velocity v making an angle 60° with the horizontal. Neglecting air resistance, the change in momentum from the departure A to its arrival at B, along the vertical directions :

Answer»

2mv
`sqrt(3)`mv
`mv`
`(mv)/sqrt(3)`

Solution :The vertical component of velocity v at the POINTS A and B in the fig. will be equal but opposite i.e. v SIN`THETA` and (- v sin`theta`).
HENCE `Deltap = mv sin 60° - (- mv sin 60°)`
`2mv=sqrt(3)/2=sqrt(3)mv`


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