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Conservation of energy using orientation |
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| Nov11-12, 11:30 AM | #18 |
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Conservation of energy using orientation |
| Nov11-12, 11:43 AM | #19 |
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| Nov11-12, 12:01 PM | #20 |
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(80 kg)(7.5 m/s [E20S] = (80kg)(V1`) + (90 kg)(3 m/s[N30E]) 600 kgm/s [E20S] = (80kg)(V1`) + 270 kgm/s [N30E] 600 kgm/s [E20S] - 270 kgm/s [N30E] 563.8155 kgm/s[E]+205.212086 kgm/s[S]+233.826 kgm/s[S]+135 kgm/s[W] = (80kg)(V1`) 428.8155 kgm/s[E] + 439.038086 kgm/s[S] = (80kg)(V1`) 613.44 kgm/s [S44.3E] = (80kg)(V1`) 7.668 m/s [S44.3E] so what i did was i multiplied the mass of player which is 80 kg by the initial velocity before the collision. The linesman is not moving so the velocity is 0m/s. And after the collision for the player we are trying to find the final velocity and so for the linesman we know the final velocity and the mass so i multiplied them. I took the linesman's momentum after collision and took it to the other side where it becomes negative. I broke the north and east compoenets Then i broke the kgm/s before and after into their separate components |
| Nov11-12, 12:14 PM | #21 |
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Looks to me like you are adding up perpendicular components like ordinary numbers. You cannot do that.
Instead write two equations: initial momentum (x components) = final momentum (x components) initial momentum (y components) = final momentum (y components) |
| Nov11-12, 01:12 PM | #22 |
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v1 = 7.5 m/s [E20S] V2`= 3m/s [N30E] x components m1v1 = m1v1` + m2v2` (80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(1.5m/s[E]) 563.8155725 kgm/s [E] - 135 kgm/s[E] = (80kg)(v1`) 428.8155725 kgm/s [E] / 80kg = v1` 5.360194656 m/s [E] = v1` y components m1v1 = m1v1` + m2v2` (80kg)(2.565151075m/s[S]) = (80kg)(v1`) + (90kg)(2.598076211m/s[N]) 205.212086kgm/s[S]-233.8268519m/s[N] = (80kg)(v1`) 439.0389379kgm/s[S]/80kg = v1` 5.487986724m/s[S] = v1` combining them: 5.360194656 m/s [E] + 5.487986724m/s[S] = v1` 7.668 m/s [S 44.3° E] = v1` And i get the same answer... |
| Nov11-12, 02:04 PM | #23 |
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OK, this time it's much clearer what you've done. (Sorry for not catching on earlier.) I don't see an obvious error in your work, so I suspect that there is a mistake in the problem statement. Double check the angles given.
(If this is from a textbook, tell me which.) |
| Nov11-12, 02:34 PM | #24 |
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| Nov11-12, 02:40 PM | #25 |
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| Nov11-12, 02:46 PM | #26 |
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| Nov14-12, 08:09 PM | #27 |
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m1= 80 kg m2 = 90 kg v1 = 7.5 m/s [ E 20 S] v1` = ? v2= 0m/s v2` = 3m/s [ E 30 N] X component m1v1 = m1v1` + m2m2` (80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E]) 563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`) 329.9887135 [E] / 80kg = v1` 4.124858919 m/s [E] = v1` y components m1v1 = m1v1` + m2v2` (80kg)(2.565151075m/s[S]) = (80kg)(v1`) + (90kg)(1.5m/s[N]) 205.212086kgm/s[S]-135m/s[N] = (80kg)(v1`) 70.212086kgm/s[S]/80kg = v1` 4.252651075m/s[S] = v1` 4.124858919 m/s [E] + 4.252651075m/s[S] = v1` 5.924 m/s [S 44° E] = v1` b) Ektot = 1/2m1v1^2 Ektot= 1/2(80kg)(7.5m/s)^2 Ektot=2250 J Ektot` = 1/2m1v1`^2 + 1/2m2v2` Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2 Ektot` =1403.75104 J + 405 kg J Ektot` = 1808.75104 J Ektot - Wdef = Ektot` 2250 J - Wdef = 1808.75104 J Wdef = 441.24896 J The Wdef is the energy lost during the collision (441.248896 J / 2250 J) x 100% 19.6 % I guess i should just re-post my answer on this thread... |
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