A projectile is fired with initial momen...

A projectile is fired with initial momentum p at an angle `45^(@)` from a point P as shown in figure. Neglecting air resistance, the magnitude of change in momentum between leaving P and arriving at Q is:

`p//2`

`psqrt(2)`

`p`

`2p`

Text Solution

Verified by Experts

The correct Answer is:

Initial momentum =p, Final momentum =-p
Angle between momentum vectors `=270^(@)`
|change in momentum|
`=sqrt(p^(2)+(-p)^(2)+2p(-p)cos 270^(@))=sqrt(2)p`

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A projectile of mass m is fired with a velocity v from point P at an angle 45^@ . Neglecting air resistance, the magnitude of the change in momentum leaving the point P and arriving at Q is

(a) `sqrt2mv`

(b) `2mv`

(d) `(mv)/(sqrt2)`

A projectile is projected with a velocity v_(0) at an angle theta with the horizontal as shown in figure. The angular momentum of particle about the origin:

is zero when particle is at the origin

is `(-mv^(3) sin^(2) theta cos theta)/(2g) hatk` when particle is at the highest point of trajectory

is `(-2mv^(2) sin^(2) theta cos theta)/g hatk` when particle is just about to hit ground

downward force of gravity exerts a torque in `-z` direction.

A 2.0-kg projectile is fired with initial velocity components v_(o x)=30 m//s and v_(o y)=40 m//s from a point on the earth's surface. Neglect any effects due to air resistance. How much work was done in firing the projectile ?

900J

2500J

1600J

4900J