Question

A pool ball rolling at 10 m/s[W] strikes a cushion and rebounds with a velocity of 10 m/s[S]. What is the change in velocity of the pool ball?

Answer 1

I initially thought that it was v2-v1, but doesn't that give you 0m/s? The answer is supposedly 14m/s.

Answer 2

W and S are the directions in which the ball rolls? If so, we can define a coordinate system with N the positive y-direction, E the positve x-direction, W negative x-direction and S negative y-direction .
The ball's velocity written as a vector before the strike then would be
\[v_i = \left( \begin{array}{c}−10 \\ 0 \end{array}\right)m/s\]
and after the strike
\[v_f = \left( \begin{array}{c} 0 \\ -10 \end{array}\right)m/s\]


The change is \[Δv=v_f−v_i = \left( \begin{array}{c} −10\\−10 \end{array}\right) m/s\]
The magnitude of the change is \(\sqrt{(−10)^2+(−10)^2} m/s \approx 14m/s\)