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5 Nov 2009 17:21:20 IST
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for the second sum simply write the co-ordinates of the particles in an arbitary point T assuming xy planne as your base reference plane and xy axis as the reference axis. then find distance between them by distance formula and equate its differential coefficent to 0. you will get a value of t. resubstitute in the distance formula equation to get minimum distance.
10 Nov 2009 01:50:08 IST
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the first one is quite simple..an application of relative motion....velocity of wedge is 10root3 i(i nd j here mean unit vectr)...nd speed of particle is 5root3 i + 15 j ...so nw velocity of particle w.r.t wedge v(pw)= -5root3 i + 15 j ....so this is same as throwing a projectile frm a base of inclind plane wid horizontal velocity 5root3 nd vertical velocity 15..u can get the time.....if u hav prob getnh it jst tell..i will post the complete ques...nd by the way the answer cums out to be 2sec.....
10 Nov 2009 01:53:06 IST
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abt the secnd one there is a simple method....use the concept of relative moton b/w projectiles...write the velocity in vectr natation of the two projectiles nd then find the relative velocity of one w.r.t other..the relativ velocity of the projectile will b along a straight line...usong simple trigo u can get the minm distance...i will soon post the complete solution....















second ques