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?
LVL 19
BrianGEFF719Asked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

ozoCommented:
I'm not sure what happens to the sand when it leaks out, but if it is moving with the same velocity as the car when it leaves the car, and then has no further interactions with the car,
I would think the velocity of 15m/s would not change.

Or maybe the sand is suposed to hit the ground and stop moving, (losing momentum as momentum is transfered to the ground)
and maybe you're then supposed to find the velocity of the centre of mass of the car+leaked sand?

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
BrianGEFF719Author Commented:
I think they want you to assume the falling sand is decreasing the mass of the system and the sand is no longer involved in the system.

brian
ozoCommented:
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
Learn Ruby Fundamentals

This course will introduce you to Ruby, as well as teach you about classes, methods, variables, data structures, loops, enumerable methods, and finishing touches.

BrianGEFF719Author Commented:
So we cant say the car is an isolated system and the mass of that system is changing?

brian
samb39Commented:
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.
ozoCommented:
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)
iloveannawCommented:
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 !!

It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.