Answer to Question #149527 in Mechanics | Relativity for Mikkos Joshua Garcia

Question

A 0.224 kg billiard ball moving at a speed of 12.0 m/s strikes the side of the table at 30° with the horizontal. The ball rebounds at the same speed and angle. What is the change in momentum of the ball?

Answer 1

The diagram above represents the vector representation of the velocity of the ball.

$\begin{aligned} \Delta v_x &= v_{iix}-v_{ix}\\ &=12sin60-12sin60\\\Delta v_x&=0\\ \end{aligned}\\ \begin{aligned} \Delta v_y &=v_{iiy}-v_{iy}\\ &=12cos60-(-12cos60)\\ &=12×0.5+ 12×0.5\\ \Delta v_y &= 12ms^{-1} \\ \Delta v &= \sqrt{(\Delta v_y)² + (\Delta v_x)²}\\ \Delta v &= \sqrt{(12ms^{-1})²}\\ \Delta v &= 12ms^{-1}\end{aligned}\\ \textsf{The change in momentum is given by,}\\ \Delta p = m\Delta v\\ \Delta p = 0.224kg × 12ms^{-1} \\ \Delta p = 2.68Ns$