[HRW7 9.P.020.] Figure 9-47 gives an overhead view of the path taken by a 0.165 kg cue ball as it bounces from the rail of a pool table. The ball's in

Wribreeminsl

Wribreeminsl

Answered question

2020-11-02

[HRW7 9.P.020.] Figure 9-47 gives an overhead view of the path taken by a 0.165 kg cue ball as it bounces from the rail of a pool table. The ball's initial speed is 2.50 m/s, and the angle 1 = 30.0°. The bounce reverses the y component of the ball's velocity but does not alter the x component. Fig. 9-47 (a) What is 2? ° (b) What is the change in the ball's linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the question.) i kgm/s + j kgm/s

Answer & Explanation

brawnyN

brawnyN

Skilled2020-11-03Added 91 answers

1.The coordinate system is drawn in a standard (2D) cross section of x right>)positive, and y up^)positive. and yes in accordance with the diagram the rebound angle of the ball does produce the same initial angle. the rebound of the ball occurs (or is drawn in the diagram) as (0,#) with the y axis. the ball starts on the left side and goes to the right side of the y-axis and would be heading away from the wall it has hit, which i may see as the negative y-direction when considering the initial + movement then hitting the wall to go - from original while the x component stays the same.
2.You haven't provided the diagram, so I can't be sure this iscorrect . . .
(a) The ball rebounds with the same angle, 30degrees.
(b) p=mv=m(vfvi)=m[(vxfi+vyfj)(vxii+vyij)]=m[(vxvx)i+(vy(vy)j]

=2mvyj=[2mvsin(θ)]j=0i+2(0.165)(2.5)sin(30)j=0i+0.4125j
The answer could be 0i+ (-0.4125)j.
Itdepends upon the coordinate system used, and without seeing thediagram, I don't know if there's an implied poisitive j direction.

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