1 / 7

Ball “A”

Ball “A” and Ball “B” have the same mass- 1.0 kg. Ball “A” is moving toward Ball “B”. Ball “B” is stationary. Ball “B”. Ball “A”. When “A” travels at 2.0 m/s and strikes “B”, Ball “B” moves upward at 35 o and Ball “A” moves downward at 55 o.

rhea-howe
Download Presentation

Ball “A”

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ball “A” and Ball “B” have the same mass- 1.0 kg Ball “A” is moving toward Ball “B”. Ball “B” is stationary Ball “B” Ball “A”

  2. When “A” travels at 2.0 m/s and strikes “B”, Ball “B” moves upward at 35o and Ball “A” moves downward at 55o . Whatever momentum “A” has before the collision A’ and B’ share after they collide rb’ ra 35o 55o ra’ mva + mvb = mva’ + mv2’ 1.0 kg x 2.0 m/s + 0 = mva’ + mvb’ 2.0 kg m/s = ra’ + rb’

  3. rB’ rB’ rA 35o 35o rA’ rA rA’ rb’ = Cos35 x 2.0 kg m/s rb’ = 1.6 kg m/s rA’ ra’ = Sin 35 x 2.0 kg m/s rB’ ra’’ = 1.1 kg m/s rA 35o

  4. rB’ rA 35o rA’ rB’ 35o rA rb’ = Cos35 x 2.0 kg m/s rA’ rb’ = 1.6 kg m/s vb’ = rb/mb vb’ = 1.6 kg m/s / 1 kg = 1.6 m/s rA’ rB’ ra’ = Sin 35 x 2.0 kg m/s rA ra’’ = 1.1 kg m/s 35o va’ = ra/ma va’ = 1.1 kg m/s / 1 kg = 1.1 m/s

  5. rBx’ = Cosq1rB’ rA = rBx’ + rAx’ rBy’ = Sinq1rB’ rBy’ rB’ rBx’ q1 rA q2 rAx’ rA’ rAy’ rAx’ = Cosq2rA’ rAy’ = Sinq2rA’ 0 = rBy’ + rAy’

  6. p = p’ mvA = mvA’Cosq2 + mvB’Cosq1 vBy’ vB’ vBx’ q1 vA q2 vAx’ vA’ vAy’

  7. Two balls of equal mass strike each other. Ball A moves at 5.20 m/s and strikes ball B, which is stationary. After the collision ball A travels above the original path at an angle of 21.00. What is the “p” and “v”s of each ball after the collision? The mass is 2.0 kg. • rA’ = Cosq1rA rA’ = Cos 21.00 x 10.40 m/s rA’ = 9.7 kg m/s • rA’ = mvA’ vA’ = 9.7 kg m/s / 2.0 kg = vA’ = 4.85 m/s 2. rB’ = Sinq1rB rB’ = Sin21.00 x 10.40 m/s rB’ = 3.73 kg m/s rB’ = mvB’ vB’ = 3.73 kg m/s / 2.0 kg = vB’ = 1.86 m/s

More Related