Can you explain how we get 8.25 from substituting 8.24 in 8.23?

College Physics
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ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
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Chapter34: Frontiers Of Physics
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Can you explain how we get 8.25 from substituting 8.24 in 8.23?
Yuzu: University Physics with
A reader.yuzu.com/reader/books/9780135206416/epubcfi/6/192[%3Bvnd.vst.idref%3DP7001015952000000000000000004273]!/4/2[P7001015
00000000000000. O *
MBUB2a = MA (VAla – VA2¤)
Now we divide Eq. (8.21) O by Eq. (8.22) O to obtain
(8.23)
V B2x = VA1x + VA2x
We substitute this expression back into Eq. (8.22) D to eliminate v B2x and then solve for v42æ:
(8.24)
mB (VAlz + VA2a)
mA (VA12 - VA2a)
mA– MB
-V Alr
mẠ+mB
V A2x
Finally, we substitute this result back into Eq. (8.23) D to obtain
(8.25)
2m A
V B2r =
VAla
mA+ mB
Now we can interpret the results. Suppose A is a Ping-Pong ball and Bis a bowling ball. Then we expect
A to bounce off after the collision with a velocity nearly equal to its original value but in the opposite
direction (Fig. 8.23a D), and we expect B's velocity to be much less. That's just what the equations predict.
1
Transcribed Image Text:Yuzu: University Physics with A reader.yuzu.com/reader/books/9780135206416/epubcfi/6/192[%3Bvnd.vst.idref%3DP7001015952000000000000000004273]!/4/2[P7001015 00000000000000. O * MBUB2a = MA (VAla – VA2¤) Now we divide Eq. (8.21) O by Eq. (8.22) O to obtain (8.23) V B2x = VA1x + VA2x We substitute this expression back into Eq. (8.22) D to eliminate v B2x and then solve for v42æ: (8.24) mB (VAlz + VA2a) mA (VA12 - VA2a) mA– MB -V Alr mẠ+mB V A2x Finally, we substitute this result back into Eq. (8.23) D to obtain (8.25) 2m A V B2r = VAla mA+ mB Now we can interpret the results. Suppose A is a Ping-Pong ball and Bis a bowling ball. Then we expect A to bounce off after the collision with a velocity nearly equal to its original value but in the opposite direction (Fig. 8.23a D), and we expect B's velocity to be much less. That's just what the equations predict. 1
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