What is the change in the y-component of the momentum (kg*m/s)? Appsa Amazon.com: Tiny. A Eligibility Requirem.JavaScript Concepts o Mail - OWENS, AYD.O Spotify R regex101: build, tes..V Vim Cheat SheetO Basic overview of M...E Reading listAyden Owens v (Student - section: 1)2021 Fall, Phys 211(Wilson)Messages Courses Help LogoutMain MenuContentsGradesCourse Contents » ... » HW 10 » cue ball with an initial speedTimerNotesEvaluateFeedback Print O Info130°(hrw8c9p20) A 0.165 kg cue ball with an initial speed of 1.15 m/s bounces off the rail in a game of pool, as shown from overhead in the figure. For x and y axes located as shown, the bouncereverses the y-component of the ball's velocity but does not alter the x-component. What is e in the figure (deg)?3.00x101 degYou are correct.Your receipt no. is 163-7997 OPrevious TriesWhat is the change in the ball's linear momentum in unit-vector notation? (The fact that the ball rolls is not relevant to either question.) What is the change in the x-component of the momentum(kg*m/s)?0.00 kg*m/sYou are correct.Your receipt no. is 163-3688 OPrevious TriesWhat is the change in the y-component of the momentum (kg*m/s)?0.3286566407kg*m/sSubmit AnswerIncorrect. Tries 5/15 Previous Tries

Dear student The momentum is defined as : P = mass × velocity Che change in momentum is given by : ∆P = Final Momentum - Initial momentumAccording to given data: The y component of initial velocity of the ball is: Vi = 1.15 Sin30° The final velocity ( after collision) is: Vf = -1.15 Sin30° Mass = m = 0.165 Kg So the change in momentum will be : ∆P = 0.165 × ( -1.15 Sin30 -1.15 Sin30) => Magnitude of change in momentum = ∆P = 0.18975 Kg.m/s