Conservation of momentum changing mass
A railroad car filled with sand has a mass 85,000kg and is moving at 15m/s. After 20min 13,000kg has leaked out what is the speed?
m1 = 85,000kg
m2 = 72,000kg
v1 = 15m/s
v2 = ?
m_1 * v_1 = m_2 * v_2
v_2 = (m_1 * v_1)/m_2
v_2 = 17.70 m/s
Is this correct, if not why?
m1 = 85,000kg
m2 = 72,000kg
v1 = 15m/s
v2 = ?
m_1 * v_1 = m_2 * v_2
v_2 = (m_1 * v_1)/m_2
v_2 = 17.70 m/s
Is this correct, if not why?
ASKER CERTIFIED SOLUTION
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Conservation of momentum only applies to a whole isolated system.
Momentum in parts of a system can change by being transferred to other parts of a system.
I still see no reason why sand leaking out of the car should affect the speed of what's left behind in this situation.
If the sand that leaks out keeps its velocity after leaving the car, then
85,000kg * 15m/s = 72,000kg * v2 + 13,000kg * 15m/s
If 13,000kg of sand lose their momentum after leaving the car
72,000kg * v2 = 85,000kg * 15m/s - 13,000kg * 15m/s
If the sand is shot backwards out of the car with enough force so that its veocity relative to the ground is 0 when it leaves the car, then
85,000kg * 15m/s = 72,000kg * v2 + 13,000kg * 0m/s
Momentum in parts of a system can change by being transferred to other parts of a system.
I still see no reason why sand leaking out of the car should affect the speed of what's left behind in this situation.
If the sand that leaks out keeps its velocity after leaving the car, then
85,000kg * 15m/s = 72,000kg * v2 + 13,000kg * 15m/s
If 13,000kg of sand lose their momentum after leaving the car
72,000kg * v2 = 85,000kg * 15m/s - 13,000kg * 15m/s
If the sand is shot backwards out of the car with enough force so that its veocity relative to the ground is 0 when it leaves the car, then
85,000kg * 15m/s = 72,000kg * v2 + 13,000kg * 0m/s
ASKER
So we cant say the car is an isolated system and the mass of that system is changing?
brian
brian
If the system is frictionless, nothing will change. Suppose the railroad car was drifting through space. You could shove out the sand, and it woud just coast along with the car, nothing would change. If you want to change the speed of the car, you need to eject the sand forcibly out like a rocket.
However, there is another way to look at this problem. Perhaps we should assume that the engine is continuously pressing forward to overcome friction, with a constant force. IF the main friction is wind resistance and does not change with the weight of the car, the speed will still remain unchanged when some sand leaks out. But if the main friction is from the weight of the car on the track, the speed will increase. However, to say how much you must model the total friction and how it changes with weight and speed.
However, there is another way to look at this problem. Perhaps we should assume that the engine is continuously pressing forward to overcome friction, with a constant force. IF the main friction is wind resistance and does not change with the weight of the car, the speed will still remain unchanged when some sand leaks out. But if the main friction is from the weight of the car on the track, the speed will increase. However, to say how much you must model the total friction and how it changes with weight and speed.
The mass of an isolated system cannot change.
(well, rest mass can change to energy, but the energy carries just as much momentum as the mass, so momentum still does not change)
(well, rest mass can change to energy, but the energy carries just as much momentum as the mass, so momentum still does not change)
was working on this long enough, so thought i'd share
m = initial mass
dm = change in mass (is negative)
dv = change in velocity
v = instant velocity of wagon
mv = (m + dm)(v + dv) + (-dm)(v + dv)
mv = mv +mdv + vdm + dmdv - vdm - dmdv
mdv = 0
but m = 85000 kg, so dv = 0 !! velocity remains constant !!
m = initial mass
dm = change in mass (is negative)
dv = change in velocity
v = instant velocity of wagon
mv = (m + dm)(v + dv) + (-dm)(v + dv)
mv = mv +mdv + vdm + dmdv - vdm - dmdv
mdv = 0
but m = 85000 kg, so dv = 0 !! velocity remains constant !!
ASKER
brian