IIT JEE , AIEEE Forum - Physics , Mechanics - A month or so back we had many of these interesting chalenges in mechanics

madman

but its been a long time since some one has put up such good challenges.

so if you know any of these good amazing mechanics questions put it up here.

BALGANESH

this is a good question

A partical of mass 1 Kg. which moves along the X-axis is subject to an accelerating forces which increases linearly with time and a retrading force which increases directly with displacement (constant of proportinatly being one with proper dimentions in both the cases). At time t=O, displacement and valocity both are zero . Find the displacement as a function of time t.

madman

keep it up guys
such questions can be of great help
so keep posting

goiit_user

Let F be the sum of all acceleration forces and P that of retarding forces
Then,
F is proportional to t and P to x with the proportionality constant equal to 1
Then,
F = t
P = x
Now,
F - P = m a
F - P = a as m=1

Displacement, x = ut + (1/2) at²x = (1/2) (F - P ) t²x = (1/2) (t - x) t²x = t³/ (t² +2)

Am I right??

BALGANESH

ans is x = t - sint (hint get an eqn and compare with the standard eqn of shm)

goiit_user

What's the problem in the answer I gave..??
See the derivation..There's no glitch

Derive how u say its t-sint

rohith291991

goiit_user u have made the common mistake of appling newtons equations of motion when a is not constant..... they can be used only when a is constant....here is m solution... the equation of motion as given in the problem is dv/dt=t-x now differentiating we get d²v/dt²=1-v or d²v/dt²=--(v-1) the solution to this kind of differential equation is of the form v-1=sin(t+c) where c is a constant...but given when t=0 v=0 hence c= -

/2 that is v=1-cost hence dx/dt= 1-cost

x=

1-cost = t-sint+d where d is a constant but given when t=0 x=0 hence d=0..hence x=t-sint

goiit_user

Hmm thats good..
Atleast now i wont make that mistake again!!
Thxx buddy

madman

nice one rohit

BALGANESH

good rohit . wanna more tough sums ?????????????????

goiit_user

ya sure buddy

rohith291991

sure...bring it on!!!

BALGANESH

will be posting soon

BALGANESH

A man can row a boat in still water at 3 Km/hr. He can walk at speed 5 Km/hr. on the shore. the water in river flows at 2 km/hr. If the man rows across the river and walks along the shore to reach the opposite point on the bank. Find the direction in which he should row the boat so that he could reach the opposite shore in the least time calculate this time.

width of the river 500 m.

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