Question

A projectile of a mass m is thrown with a velocity v with an angle 60 with the horizontal. Nelecting air resistance ,the change in momntum from the depature A to its arrival at B, along the vertical direction is​

Answer 1

Answer:

Change in momentum...

2mvsin 60

2 mv √3/2

√3 mv

Hope it helps!!!

Answer 2

Answer:

change in momentum$=\ \sqrt{3}.m.v$

Explanation:

Given,

mass of projectile = m

angle made with horizontal= 60

Let's consider the velocity of projectile when thrown from point A is v then the velocity of projectile at B, when fall down is -v.

so, momentum of projectile at A= m.v sin(60)

                                                     $=\dfrac{\sqrt{3} }{2} \times m.v$

so, momentum of projectile at A= m.v sin(60)

                                                     $=\dfrac{-\sqrt{3} }{2} \times m.v$

change in momentum = initial momentum - final momentum

                                     $=(\dfrac{\sqrt{3} }{2}.m.v)- (\dfrac{-\sqrt{3} }{2} .m.v)\\\\=\sqrt{3}.m.v$