IIT JEE , AIEEE Forum - Physics , Mechanics

vasu314

A train of mass M is moving with a uniform velocity and suddenly a wagon of mass m gets detached. The driver of the engine realises this after travelling a distance L and shuts off the engine. Assuming the power of the engine to be a constant, and the resistance is proportional to weight, find the separation between the train and the wagon when both of them come to a halt.

elessar_iitkgp

If its pull instead of power, then learner's approach is correct. But, as vasu said, it is much more interesting if power is constant.

vasu314

Yeah,.. u need to take the assumption into account.

The answer is ML/(M-m)

..

learner

dude this isnt a question on power-its pure kinematics.

I'd asked this question earlier but couldn't get the correct answer.Here goes:

Initial velocity=v...say

Pull of engine=resistance(no acc.)=KM=>K=a

When uncoupled,extra pull=KM-K(M-m)

=Km

Acc of train=Km/(M-m)

Retardation for bogey=K

Distance travelled by train=L

& L=vt+0.5(Kmt^2/(M-m))....2nd eqn of motion..(remember this relation)

=>2L/t=2v+Kmt/(M-m)

Once steam shuts off=s..say

V^2-U^2=2as

It stops,so V=0

So,U^2=2Ks

2Ks=(v+Kmt/(M-m))^2.....1st eq of motion

=>s=[(v+kmt/(M-m))^2]/2K....remember this

Distance travelled by last bogey=s2...say

s2=v^2/2k....3rd eqn of motion

Distance between them=l+s-s2

Simplify it...its slightly tricky and you'll get your answer..

Thats quite some typing..

Rate if it helped!!

vasu314

Good attempt..The sol'n was on expected lines...but i think the assumption that the acceleration provided by the extra pull K*m is a constant is not justified.

The reason is as follows.

The extra pull just after the separation is as rightly pointed out is K*m. Hence the body will accelerate. The moment it starts accelerating, its velocity increases and hence the driving force, which is used to overcome the resistance and provide the extra pull, decreases. This is so because the power is a constant. P= F*v

As the driving force reduces, so does the extra pull and hence acceleration is decreasing with time. Therefore, the simple kinematic equations are invalid.

learner

Nice deduction...I didnt think of it earlier...

.

But in the place I read the question when I posted it, it said pull,not power is constant...

So pull= M*a would imply constant acceleration..

Taking that as the basis,I proceeded with the rest of the problem.

That being the question,what I did would be correct and you get Ml/(M-m)..

.

Then again,as you pointed out,the same principle cant be applied here,which falsifies that solution..

So I'm unable to understand how the answer remains the same in both cases..

someone please explain!!

@Elessar sir, your post is slightly unclear as many symbols cant be seen..please check it!

Thanks!:-)

vasu314

Yeah..

So was i surprised when i saw that this was the answer given..I got an answer in an exponential form...So perhaps, it is pull that was intended...

But it makes for a nice problem when Power is a constant....

learner

i'll try it that way...it seems complicated though!!

Nice question!

vasu314

Learner...

could u suggest from which book this problem on power is present...

learner

I got it from a question bank from my library...by Prof.EV something if I remember correctly...just happened to remember it since I too had asked it once.

vasu314

Wud be glad if u cud help me by finding out the title of the book and its author..

Thnx a lot...

Cud u also suggest any other gud books which contain challenging questions in Physics in general,..and optics, electricity and modern physics in particular...

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