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

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?

Experts Exchange Solution brought to you by

Enjoy your complimentary solution view.

Get this solution by purchasing an Individual license!
Start your 7-day free trial.

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.

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 trialbrian

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

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.

(well, rest mass can change to energy, but the energy carries just as much momentum as the mass, so momentum still does not change)

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 !!

Math / Science

From novice to tech pro — start learning today.

Experts Exchange Solution brought to you by

Enjoy your complimentary solution view.

Get this solution by purchasing an Individual license!
Start your 7-day free trial.

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?