Messages posted by yahiyafirdous

Let us make a few assumptions before solving the problem as stated below:

1. After collision the block moves along the surface of wedge with respect to it.

2. The wedge moves in horizontal direction.

Let us assume x-aixis along the surface of wedge and y-axis along normal to the surface of wedge.

From FBD of block:

Applying conservation of momentum along x-axis:

....................................(1)

Along y-axis:

....................................(2)

From FBD of wedge:

.........................................................................(3)

Solving above equations for V12 we get

:

Once relative velocity of block with respect to wedge is found at t=0, now we will try to find Relative acceleration of block wrt wedge.

See next part of solution.

dy/dx = 2/y .................(1)

r=(1-x)i -y j

Putting all values r , dl etc we get

Integrating dB , x=0 , x= infinity and multiplying the result by 2 we get

tan@=y/x

x=yCot@

a=gCot@ Answer

now put the value r0

now put the value r0

now put the value r0

Let a1 is acceleration of block-1 and block-3 and a2 is acceleration of block-2.

From FBD:

Solving above equations for a1 and a2 we get:

Distance travelled by Block-1:

Distance travelled by Block-2:

But

Hence

Sorry I foregot that mass of B is M, just replace m by M of Vbx in equation (2) and solve

Let they are placed r meter apart,

F=kq1(q-q1)/r^2

dF/dq1= k(q-2q1)/r^2 =0 for F maximum

Hence q1/q= 0.5 Ans

Let the velocity of sphere A is V0 before impact , Vax and Vay are the X and Y components of Sphere-A after impact. Vbx is velocity of Sphere B after ipmact. Since there is no impulsive force along Y-axis, hence Velocity in Y-direction is not affected at all(Since balls are smooth).

Apllying conservation of Linear momentum along x and y axis:

............................(1)

.........(2)

......................(3)

Solve above equations for Vbx

we get Ans

Fundamental mistake by Anitha is "Conservation of Energy" Here its an inelastic collision, hence energy is niever conserved. Secon mistake is the assumtion that verticle component of velocity is zero. If verticle component is zero how the smaller block will reach the height h.

INFORMATION IS INCOMPLETE. FIRST WHETHER TUBE IS IN VERTICLE PLANE, WHAT IS R. PLEASE ENCLOSE FIGURE.

Mr Anirban is my answer is correct?? I hope it must be correct.

Let us apply Kirchoff's loop(current) law and assume all current is flowing from positive to negative terminals. Since C1 and C4 are parallel to voltage source , hence voltave across them are 20V and 10V respectively.

Since current through C2 and C3 are equal hence:

.........................(1)

..................(2)

Solving above equations

Vc3=3V

Vc2=7V

Vab=Vc4+Vc2= 17V Ans

I solved this problem and is a little lengthy. I am getting

I will send you complete solution later on.

Your answer is

This is easy to solve. Let the initial and final position of COM of the system from pulley are Yc0 and Yc1 which can be found to be :

Applying conservation of energy:

Solving above above equation for v:

we get:

Let velocity at when it is moving verticle is v,

.......................................(1)

Hence

Normal component of accn is given by

Tangential Component= g

Hence resultant accn is

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