Timeline for Conservation of momentum when rain pours into a wagon
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 10 at 15:26 | comment | added | Amit | Brilliant explanation. However, in the vertical direction, taking the system to be the rain and wagon, momentum is not conserved: y momentum before = mass of rain x y component of rain and momentum after = mass of rain x zero because the y velocity of the rain after falling into the wagon is zero. The wagon has no y component of velocity so it doesn't feature in this equation. So how is it that momentum is not conserved when you take the system to be the rain and wagon in the y direction? | |
| Dec 15, 2014 at 22:44 | vote | accept | math_lover | ||
| Dec 15, 2014 at 22:44 | vote | accept | math_lover | ||
| Dec 15, 2014 at 22:44 | |||||
| Dec 15, 2014 at 16:37 | comment | added | thomij | +1 for focusing on the definition of the system - the most common source of confusion in classical physics! | |
| Dec 15, 2014 at 13:34 | history | edited | DJohnM | CC BY-SA 3.0 |
added 1 character in body
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| Dec 15, 2014 at 6:29 | history | answered | DJohnM | CC BY-SA 3.0 |