# Homework Help: Variable mass type question

1. Aug 11, 2014

### Sagar98

A cart has a load of 100 kg sand at start and it is pulled by a steady resultant force of 100N starting from rest. A hole in the cart causes steady leakage of 1kg/s. Find velocity v of the cart after 50s of motion.

dF =(dM /dT) *(dV/dT)
dM/dT, since the mass varies with time and dV /dT that us the acceleration. I tried my best. Now someone correct my logic if wrong and tell me the concept to do this. Thanks in advance. I've posted this question like 3-4 times and someone deleted it for random reasons.

2. Aug 11, 2014

### Staff: Mentor

It was not deleted for random reasons. Check your PMs -- you will find the text describing your rules violations there.

3. Aug 11, 2014

### stateofdogma

They probably deleted it because you didn't follow the rules for posting homework problems https://www.physicsforums.com/showthread.php?t=686781
Rearrange this equation,
$$F = \frac{d(v)(\triangle M)}{dt}$$
for a momentum equation and take the integral of both sides of the equation with respect to time ,
Pay attention to the DeltaMass, then the rest is simple integration.

Last edited: Aug 11, 2014
4. Aug 12, 2014

### Sagar98

I got it, It's like this,
t-time taken
v-velocity
a-acceleration
Now,
(100-t)×a=100
That is ma=Force. But 100-t,as in that the mass reduces as the time increases.
=> (100-t)dv/dt =100
=>dv=100/(100-t) dt
Now substitute 100-t as u, so dt =-du.
Therefore
=> v=-100 integral du/u
=> v=-100 ln (u)
=>v=-100 ln(100-t)
=>put limits from 0-50 secs
v=100(ln100-ln50)
=> v=100(ln(2))
v=69.3m/s

5. Aug 12, 2014

### Sagar98

@state of dogma, thanks..

Last edited by a moderator: Aug 12, 2014
6. Aug 13, 2014

### dean barry

Cant you use average acceleration rate ?
At the beginning its = 100 / 100 = 1 (m/s)/s
After 50 seconds its = 100 / 50 = 2 (m/s)/s

Average acceleration rate (a) = 1.5 (m/s)/s

Velocity after 50 seconds = a * t = 75 m/s

7. Aug 13, 2014

### dean barry

Sorry, just done it using increments on excel, i got 69.314 m/s

8. Aug 13, 2014

### Sagar98

Yep correct ^ I did it by writing so calculations ain't as accurate.

9. Aug 13, 2014

### Sagar98

Your method is wrong @dean Barry, the First post - The masses is constantly changing with time, so we don't know the velocity or the acceleration at 50s, So mass is a function of time here.

10. Aug 13, 2014

### Simon Bridge

That's a reasonable bit of numerical work - but aren't you expected to use calculus to solve this problem? The trick is to use the definition of Force (Newton's second Law) as a differential equation.
Let me start you off... $$F=\frac{dp}{dt}: p(t)=m(t)v(t)$$ ... you need an expression for m(t) and the chain-rule. Set it up as an initial value problem.

11. Aug 13, 2014

### haruspex

I don't think that's going to work. F is constant here, so you can write down Ft = etc., but in that equation m has to include the lost sand, so you have to figure out the momentum that carried away with it.

Sagar98, what mass of sand is in the cart at time t? Assuming this includes the mass of the cart, how fast will it be accelerating at time t?

12. Aug 13, 2014

### Simon Bridge

dp=Fdt won't work, no; it gets: mv=Ft => gets v=100/(100-t) which gives v(0)=1 which is false.

I was trying to figure out how the person setting up the question expected them to do it.
I had a go using F=m(t). dv/dt and got the same answer as the numerical one above.

Using F=d(mv)/dt + chain rule and equation for m(t) gives me a solvable DE but I didn't take it further.
Normally you can't use it - but normally the mass lost contributes to the force so the force is not a constant.

Checking - hmmm: the solution does not make sense for the problem... thinking about it - it was probably too much to hope that the problem was that well set up.

Maybe go back to F=ma, same advise only the chain rule is no longer needed?

13. Aug 13, 2014

### Sagar98

Mass of the cart is negligible, It's all the mass of sand.

14. Aug 13, 2014

### Sagar98

@Simon Bridge, Refer to my solution , it's correct.

15. Aug 13, 2014

### haruspex

Quite so - I must have overlooked post #4 previously. Well done.

16. Aug 13, 2014

### Simon Bridge

Yeah - you did it by F=m(t)dv/dt. Sorry I thought you had evaluated it numerically too ... 2am effect.

Technically F=dp/dt ... so what assumption were you making to simplify the calculation?
(Since it was "correct", this was the assumption you were supposed to make.)

$$F=\frac{d}{dt}(mv)\\ \implies 100=(100-t)\frac{dv}{dt} -v\\ \implies v(t)=\frac{c+100t}{100-t}$$

$v(0)=0\implies c=0 \implies v(50)= 100\text{m/s}$ ... So which is physically correct?

However, since you got the assignment answer correct, you may prefer to move on ;)

(Note: I messed up with some of my comments prev post.)

17. Aug 13, 2014

### Simon Bridge

... that would make the answer non-physical for times approaching t=100s ;)
How is the equation changed if the mass is not negligible?

18. Aug 14, 2014

### Sagar98

@Simon Bridge, good question, at t=100s,There is No mass and no acceleration on it... So maybe.. It just goes at the velocity that it gained during first 100 seconds

19. Aug 14, 2014

### ehild

@Simon Bridge: see http://en.wikipedia.org/wiki/Variable-mass_system

The whole momentum of the cart and sand system does not change when some sand is spilt out, as the sand has the same velocity as the cart, (the spilt sand goes only outside the cart) but the force does not accelerate the spilt sand.

So the net momentum change from time t to t+Δt is p(t+Δt) - p(t) = [(M(t)-Δm)(v+Δv) + vΔm]-M(t)v=M(t)Δv + ΔmΔv, but the second term is negligible with the limit Δt→0.

So dp/dt=M(t)dv/dt=F for the whole cart-sand system.

If M(t)=Mo-ct, the solution is v-v(0)=F/c ln(Mo/(Mo-ct)). In this problem, it is v=100ln(100/(100-t)). The cart has nonzero mass, so the denominator is never zero. The speed changes up to time tmax=100 s. At t=50 s, the speed is 100 ln(2), the value Sagar98 got.

20. Aug 14, 2014

### ehild

You have to include the integration constant: v=-100 ln(100-t)+C

ehild

21. Aug 14, 2014

### Sagar98

Yeah, but in definite integration with limits you don't put the constant c..

22. Aug 14, 2014

### ehild

then you have to write the solution of the differential equation including the initial condition. So v(t) =v(0)-100 ln(100-t)+100ln(100) =v(0)+ ln(100/(100-t))

If the question was "find the mass when v=10 m/s", what would you get with your formula v=-ln(100-t)?

ehild