Vector components - change in momentum

[studypersist]

Homework Statement

An overhead view of the path taken by a ball of mass m as it bounces from the rail of a pool table. The ball's initial speed is v and the angle is a. The bounce reverses the y component of the ball's velocity but does not alter the x component. ... (b) What is the change in the ball's linear momentum in unit - vector notation?

Homework Equations

Change in momentum (DP)
DP(J) = mvcos(a)(-j^) - mvcos(a))+j^)
DP(I) = mvsin(a)(-i^) - mvsin(a))+i^)

The Attempt at a Solution

I know that the initial and final angles are the same. I thought I needed to use a form like this to get the answer

DP(I) + DP (J)

Both of my answers are negative. I'm not sure if that's the problem or if I have my sin and cos for j and i mixed up.

 

[Hootenanny]

If angle a is measured in the normal way (anti-clockwise from horizontal) then yes, you have your sines and cosines the wrong way round. You can verify this yourself by drawing a right-angled triangle with the momentum vector as the hypotenuse and the components as the adjacent and opposite sides.

If you know that the x-component remains unchanged, then there is no need to consider it in your calculations since motion in the y-direction is independent of motion in the x-direction. Instead, just consider the change in the y-component of momentum.