Question

A particle of mass m is projected
with velocity v making an angle of
45" with the horizontal. When the
particle lands on the level ground
the magnitude of the change in its
momentum will be:
(1) muroot2
(2) Zero
(3) 2mu
​

Answer 1

Answer:

Initial momentum=m(vcos45i+vsin45j)

Final momentum=m(vcos45i-vsin45j)

change in momentum=-2mvsin45j

Magnitude of it is 2mvsin45 or √2mv.