Question

A particle of mass m strikes elastically on a wall with velocity v, at an angle of 60° from the wall then magnitude of
change in momentum of ball along the wall is
(1)
Zero
(2) 2mv
(3)
3 mv
(4) mv​

Answer 1

Answer:4nd option - mv

Explanation:

Change in momentum ∆p=p2 - p1

=2mvcos60

=2*m*v*1/2

=mv

Answer 2

Answer:

4 mv

Explanation:

see components of velocity

are

vsin60°=√3v/2

vcos60°=v/2

since in elastic collision momentum is conserved so

px=mv/2

py=√3mv/2

p=√((mv/2)^2+(√3mv/2)^2)

p=1/2*mv√4

p=mv