1where is te figurea?
Two blocks A and B are attached to a spring of force constant K and are placed on a horizontal surface. The coefficient of friction between block A and surface is zero while between block B and surface is μ. A spherical ball of mass M, radius R impinges on block A with angular velocity w and linear velocity u as shown.
The coefficient of restitution is 1/2 and coefficient of friction between the ball and block A is 1/√2. Find velocity of block B when spring suffers maximum deformation. Initially the spring is non deformed. (M = 1kg, Ma=3kg, u=10m/s, μ=0.2, Mb=10kg, k=300/16 N/m)
NOTE- TIME ALLOTTED IS 3 MINUTES
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13 Answers
1dude all i could find out was that after collision the block a has velocity u/4 towards right...age ka part if u cn post the solution plz do it!
1law of restitution and impulse momentum theorem[1].tht kills the first part of the problem that is finding velocoity after collision....just after collision i am not able to think what approach to take..[2](the friction between block B and ground is creating trouble.....oh wait...i guess i got the second part!...let me think for a bit more!)
1the velocity of block B at maximum compression is coming zero
is it correct??
i m asking kyunki yadi galat answer hua toh wats the use of typing wrong solution :P:P
1law of restitution and impulse momentum theorem
I think you misunderstood the problem. Can you write the equations for both law of restitution and impulse-momentum theorem so that I can see if any thing is wrong (or perhaps correct myself)?
No 0 wrong hai.......plz write your equations upto finding out the velocity.
71This might serve of some use : http://www.targetiit.com/iit-jee-forum/posts/rotation-instantaneous-centre-of-rotation-12434.html
262using e concept and conservation of momentum we get Va=5/3√2.. but what after that?? now there are 2 unknowns- extension in spring as well as velocity of b.
1kunl_8072: he m telling here
0:30
kunl_8072: 1.v=1/2*u/root2
0:31
kunl_8072: now using momentum impulse theorem
0:31
kunl_8072: we get impulse of normal reax
0:32
kunl_8072: and hence tt of ftiction as 3/4u
0:32
kunl_8072: now velocity of block a =3/4*3 u
0:32
kunl_8072: 3 and 3
0:32
kunl_8072: cncel
0:33
kunl_8072: so
kunl_8072: u/4kunl_8072: he m telling here
0:30
kunl_8072: 1.v=1/2*u/root2
0:31
kunl_8072: now using momentum impulse theorem
0:31
kunl_8072: we get impulse of normal reax
0:32
kunl_8072: and hence tt of ftiction as 3/4u
0:32
kunl_8072: now velocity of block a =3/4*3 u
0:32
kunl_8072: 3 and 3
0:32
kunl_8072: cncel
0:33
kunl_8072: so
kunl_8072: u/4
1now further to move block f=kx
which gives x=16*0.2/3
now for tht vibrational energy must be 1/2kx2=10.66
but the energy of block A is 1/2mv2=9.375......so i m still sure of my answer at max. compression B has velocity=0!
1this question is from BT IIT FULL syllabus test series 2 paper 2 physics
Since I gave it online I don't have access to solution. If anyone else on TIIT gave it in offline mode please post the answer here.
