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.

 

[Moetasim]

I would first clarify that the homework problem seems to be describing the physical concept of conservation of momentum. In this scenario, the ball's momentum is changing due to the bounce, but the total momentum of the ball and the pool table remains constant.

To answer the question, the change in momentum can be represented in unit-vector notation as DP = m(vcos(a)i^ - vcos(a)i^) + m(vsin(a)j^ - (-vsin(a))j^). This simplifies to DP = 2mvsin(a)j^, indicating that the change in momentum only occurs in the vertical direction (j^) and is dependent on the mass of the ball, its initial speed, and the angle of the bounce. The negative sign indicates that the ball's momentum is in the opposite direction after the bounce.