Because the collision is elastic it will bounce back with the same speed, having the same direction. The change in total momentum of the ball, along the hitting direction, is simply

`Delta(P) = m*v -(-m*v) =2*m*v`

By definition the angular momentum of a mass having the linear momentum `P` with...

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Because the collision is elastic it will bounce back with the same speed, having the same direction. The change in total momentum of the ball, along the hitting direction, is simply

`Delta(P) = m*v -(-m*v) =2*m*v`

By definition the angular momentum of a mass having the linear momentum `P` with respect to a certain point is

`L = r xx P`

where all the quantities are vectors and r is the distance from linear momentum application to the given point. Thus the variation of angular momentum in the figure is just

`|Delta(L)| = |r xx Delta(P)| = |r xx 2mv| = d*2mv*sin(pi/2 -theta) =2mvd*cos(theta)`

**The correct answer is B) `2mvd*cos(theta)` **

**Further Reading**