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							Answer is 5ucos(theeta)/2.Initial and final momentum along y direction is zeroApplying conservation of momentum along x direction just before and just after explosion :4mucos(theeta) = -mucos(theeta) + 2mv=> v = 5ucos(theeta)/2Note that the whole mass of particle before explosion is 4m and at the highest point of it`s trajectory, it has only x-component of velocity I.e. vcos(theeta)So initial momentum along x direction is (4m)ucos(theeta). ( Note that we have assumed right direction as +ve and left direction as negative)For final momentum of system, the particle of mass m which retraces it`s path will have velocity of -ucos(theeta) ( negative sign because it retraces it`s path, i.e. it`s final velocity is along left direction).Further let the velocity of 2m be v.So final momentum along x-axis is -mucos(theeta) + 2mvNow equating initial and final momentum along x-axis4mucos(theeta) = -mucos(theeta) + 2mv=> v = 5mucos(theeta)/2.So final velocity of 2m is 5mucos(theeta)/2 along right direction.