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Mechanics

Blazing goIITian

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27 Feb 2008 17:14:59 IST

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1447

Variable mass tough one....rates assured

None

A cart of total mass M is at rest on a rough horizontal road. It ejects bullets at a rate of lambda kg/s at an angle heta with the horizontal and at a velocity u(constant) relative to the cart. The coefficeint of friction between the cart and the ground is . Find the velocity of the cart as a function of t. The cart moves with sliding.

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Anand Hegde

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2 Mar 2008 09:25:25 IST

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Both the answers are wrong......I have given the answer.....There seems to be a slight mistake in your answer

abhishek sinha

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Posts: 926

2 Mar 2008 12:10:28 IST

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It is a simple problem of variable mass ( Here we take the symbol of as @ for ease of typing )

See Resnick Halliday for the Key eqn to be used i.e.

mdv/dt = Fext + v(rel ) dm/dt ......................... ( 1 )

Now considering only horizontal component , we have

m = Mo - t

dm/dt =-

v(rel ) =- u cos @

Fext = - ( mg + u sin @ )

So the above eqn becomes

mdv/dt = u cos @ - (mg + u sin @ )

or , dv = {u ( cos @ - sin @ )/ ( Mo - t ) - g }dt ( as , m= Mo - t )

integraing we get ,

v = ( u cos @ - sin @ ) ln Mo/ (Mo -t ) - gt ( Proved )

abhishek sinha

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2 Mar 2008 12:26:15 IST

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As regards to the ans that anandghegde has given , there should be a mu instead of lambda in the first bracket , otherwise the given expn is dimensionally wrong !!

anchit saini

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2 Mar 2008 12:30:30 IST

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yes on rechecking my answer (with help from feynmann) and editing(edited part in red) i am also getting the ans given by feynmann

sinjan jana

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4 Mar 2008 22:34:06 IST

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I dunno whether my ans is correct or not

The external force is the force given by the recoiling momentum.
F(ext)=(dM/dt)ucos(theta)
The normal force is M(t)g-(dM/dt)usin(theta)
tThe frictional force is -friction coefficient*normal force.
Therefore the equation of motion is
d(M(t)v(t))/dt=F(ext)+frictional force.
If dM/dt=-lamda, then M(t)=Mo-lamda*t
Solving the above two equations we can get v(t)

sinjan jana

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5 Mar 2008 10:32:33 IST

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ive cracked it
forces in y direction
1)upward normal :N
2)downward weight :(m-?t)g
3) downward impulse : ?usin@

forces in x direction
1)towards one left impulse : ?ucos@
2)towards right friction : µN

equations in y
N=(m-?t)g+?usin@

equations in x
?ucos@-µN=(m-?t).dv/dt (note variable accn)

put N from above equation
and integrate correctly u'll get the correct ans

after substituting N

[?ucos@-µmg-µ?usin@]/(m-?t) +(µ?gt)/(m-?t)=dv/dt

note ive separated the equation in two parts so as to integrate easily

thus

v= ?[?ucos@-µmg-µ?usin@]/(m-?t) dt+?(µ?gt)/(m-?t) dt

limit 0 to t

integrate correctly
u'll get the correct ans

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