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Vector components - change in momentum

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

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

    3. 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.
  2. jcsd
  3. Mar 7, 2008 #2


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    Gold Member

    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.
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