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This question is really tough for me

This question is really tough for me

by Sinjan Jana -
Number of replies: 7
 
PLEASE GIVE ME THE STEPS

A cart of total mass M is at rest on a rough horizontal road. It ejects bullets at a

 rate of   \(\lambda\frac{kg}{s}\)   at an angle  \(\theta\)  with the horizontal

 and at a velocity u(constant) relative to the cart. The coefficient of friction

 between the cart and the ground is  \(\mu\) . Find the velocity of the cart as

 a function of t. The cart moves with sliding.


THE ANSWER IS

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In reply to Sinjan Jana

Re: This question is really tough for me

by lokesh sardana -

at any time t,

normal rxn on cart =  &usin@ +  (M- &t)g

Fnett on cart = u&cos@ - \mu{(M-&t)g + &usin@}

(M-&t).dV/dt = u&cos@ - \mu{(M-&t)g + &usin@}

dV = {u&(cos@ - \musin@)/(M-&t)- \mug}dt

integrating both sides

V = u&(cos@ - \musin@)ln{(M- &t)/M} - \mugt

where & = \lambda\frac{kg}{s} and  @ = \theta

this is the answer acc. to me...slightly diff from urs..

may be I am wrong...

In reply to lokesh sardana

Re: This question is really tough for me

by Sinjan Jana -
No, you are absolutely correct.. The answer I gave is dimensionally incorrect..!!
In reply to lokesh sardana

Re: This question is really tough for me

by Chandni Bhatia -
Can you please explain why &usin@ is added to normal rxn.?
In reply to Chandni Bhatia

Re: This question is really tough for me

by lokesh sardana -
because the canon is at some angle.......so, force exerted by it will be resolved in two components, horizontal component will push the cart forward and vertically component will try to push the cart downward thus adding up in normal rxn..
In reply to lokesh sardana

Re: This question is really tough for me

by lokesh sardana -

  truely saying, my above method is not correct....

here, I first calculated Fnett and then equated it to m.a = m.dV/dt

but this is not valid because m is also changing with time..

so, it should be equal to d(mV)/dt.

but by this method , calculation become more complex and asnwer coming out to be completely different...

so, I am also confused..

sorry !!!! :(

It is better that experts should enter and tackle this problem.

I myself waiting for their replies.

Lokesh

In reply to lokesh sardana

Re: This question is really tough for me

by Sinjan Jana -
The answer given is wrong.

And the correct answer is :

(ucosθ-µusinθ)ln M/(M-λt)-µgt

we can easily see that the answer I posted is dimensionally incorrect.

The following gives a gist of the solution that is sufficient for us to understand.

Initially forget about the friction, just calculate the thrust/force generated due to recoil.

Apply Law of conservation of Momentum
In x direction,
λdt u cosθ = (M- λt)(dv) dv =change in velocity of truck.

Dv/dt= λ u cosθ/(M- λt)

Now consider frictional force = µN

N = mg + Thrust in downward direction due to recoil.

Nett retardation = dv/dt = -µ(g +( λ u sin θ/(M- λt)))

Nett acceleration = dv/dt = (λ u cosθ)/(M- λt) - (µλ u sin θ)/(M- λt) - µg

Integrate this (limits 0 to t)to get the following answer:

(ucosθ-µusinθ)ln M/(M-λt)-µgt
In reply to Sinjan Jana

Re: This question is really tough for me

by lokesh sardana -

well, this seems to be another form of my previous solution....

and I m sorry for slight mistake in my answer .....an extra '&" has been additionally multiplied in my answer due to calculation mistake in integration..

here u have written acc. a = F/m

 but is in this case newton's law applicable???? because here mass is also changing with time...

so,  it seems that same mistake has been performed as I did

that's why I m confused........