Change in momentum when given the speed (not the velocity).

[MattDutra123]

1. The problem statement.
A tennis ball of mass m moving horizontally with speed u strikes a vertical tennis racket. The ball bounces back with horizontal speed v.

Homework Equations

p = mv

The Attempt at a Solution

My answer was m(v-u), meaning the final momentum (mv) subtracted from the initial momentum (mu). It turns out the answer is m(u+v), with the justification being that we are given the speed, not the velocity. I don't see how that changes anything. How can you find the change (difference) between two quantities without subtracting them? I reckon this is a very basic question, but I don't understand it.

 

[kuruman]

You are looking for the magnitude of the change in momentum. Say the ball is moving to the right. After bouncing off the racket it's moving to the left and has negative velocity. The change in momentum vector is Δp = m (-u) - m v = -m (u+v). The magnitude of that is
+m (u + v ). If the ball is moving to the left and bounces to the right, Δp = m u - m (-v) that has magnitude m (u + v ), same thing.

 

[MattDutra123]

You are looking for the magnitude of the change in momentum. Say the ball is moving to the right. After bouncing off the racket it's moving to the left and has negative velocity. The change in momentum vector is Δp = m (-u) - m v = -m (u+v). The magnitude of that is
+m (u + v ). If the ball is moving to the left and bounces to the right, Δp = m u - m (-v) that has magnitude m (u + v ), same thing.

Thanks for the help.