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Mechanics
27 Feb 2008 17:14:59 IST
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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 
kg/s at an angle 
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. 
Comments (25)
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2 Mar 2008 03:00:58 IST
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@ Anandghegde,
Before solving this problem just go through this link;
http://ocw.mit.edu/NR/rdonlyres/Aeronautics-and-Astronautics/16-07Fall-2004/EDABC2D9-6030-4F56-B369-D6CEE666F502/0/d27.pdf
And make sure that you are perfect with all the concepts in it.
Let us proceed,
As we know,
Given, Lambda (L) = dm/dt
As, Fth = (dm/dt) u
Thus, we have,
Fth = Lu .. (i)
The direction of Fth will be in the opposite direction of u, because the system is loosing mass.
We also have variable mass = M - Lt .. (ii)
Draw the FBD and resolve Fth along x and y coordinates as shown in the Figure.
Along y coordinate, we have,
N = Fth sin 0 + (variable mass) g
Substitute (i) and (ii), we have,
N = Lu sin 0 + (M - Lt) g ..(iii)
Frictional force f = µ N ..(iv)
Along x coordinate, we have,
Fth cos 0 - f = (dv/dt)(variable mass)g
Substitute (i), (ii), (iii) and (iv), we have
Lu cos 0 - µ [Lu sin 0 + (M - Lt) g] = (dv/dt)(M - Lt) g
Multiply throughout by dt and and divide by (M - Lt)
And then integrate LHS from zero --> t and RHS from zero --> v
The sign
0t{ Lu cos 0 dt / (M - Lt) - 0t{ µLu sin 0 dt / (M - Lt) + 0t{ µg dt
Therefore, we have
v = Lu cos 0 ln | M / (M - Lt) | - µLu sin 0 ln | M / (M - Lt) | - µgt
Take ln | M / (M - Lt) | common and you will get the final answer.
I believe that whatever I have solved is correct. Nudge me if you have any query.
Dude, you are in XIth with me; you should have tried it yourself. Nevermind !!!
2 Mar 2008 08:22:40 IST
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3 people liked this
here is my attempt, i don't know why the answer is not coming but nevertheless--
i am going purely by the figure, rather the hint, that has been given now--
N= Lu sin Q + (M-Lt)g
in horizontal direction
Lu cos Q -
N=ma=(M-Lt)a=(M-Lt)dv/dt
Lu cos Q -
Lusin Q -(M-Lt)g= (M-Lt)dv/dt
dv=Lu cos Qdt/(M-Lt) -Lusin Qdt/(M-Lt) - gdt
integrating from 0 to v on l.h.s and 0 to t on r.h.s
v=(Lu cos Q -Lu sin Q) ln[M/M-Lt]/L - gt
so v=(u cos Q -
usin Q) ln[M/M-Lt] - gt
i am going purely by the figure, rather the hint, that has been given now--
N= Lu sin Q + (M-Lt)g
in horizontal direction
Lu cos Q -
N=ma=(M-Lt)a=(M-Lt)dv/dtLu cos Q -
Lusin Q -(M-Lt)g= (M-Lt)dv/dtdv=Lu cos Qdt/(M-Lt) -
Lusin Qdt/(M-Lt) - gdtintegrating from 0 to v on l.h.s and 0 to t on r.h.s
v=(Lu cos Q -
Lu sin Q) ln[M/M-Lt]/L - gtso v=(u cos Q -
usin Q) ln[M/M-Lt] - gtFree Sign Up!
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i cannot solve this
sorry