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
X Zero
(2) 2my
:) 13mv
(4) mv​

Answer 1

Answer:

$\Delta P=mv$

Explanation:

It is given that,

A particle of mass m strikes elastically on a wall with velocity v, at an angle of 60° from the wall,

Let $P_1\ and\ P_2$ are the initial and final momentum of the ball. As momentum is a vector quantity. So,

$\Delta P=\sqrt{(mv)^2+(mv)^2-2\times mv\times mv cos(60)}$

$\Delta P=mv$

So, the change in momentum of ball along the wall is mv. Hence, the correct option is (4).