A projectile is fired with velocity u at an angle @ with horizontal.At the hi e total mass of the particle = 4minitial velocity= uangle with horizontal=@vertical component of velocity=usin(@)horizontal component of velocity=ucos(@)let velocity at the highest point= vand the velocity of fragment of mass 2m =wat the highest point,there is no vertical velocityi.e v= horizontal component of velocity=ucos(@)(constant, as there is no external accelation in horizontal)at the highest poit of trajectory, net force in horizontal direction is zerohence momentum in horizontal direction is conservedi.eP(just before explosion)=P(just after explosion)=> 4m*ucos(@)=m*0+m*(-ucos(@))+2m*w=>5ucos(@)=2w=> w=(5/2)ucos(@)Ans.my doubt is why we have not taken 2mvcosghst point of its trajectory it splits up into 3 segments of masses m,m, and 2m.First part falls vertically downward with zero initial velocity and second part returns via same path to the point of projection.The velocity of third part ofmass 2m just after explosion will be what???

nidhi , 7 Years ago

Grade 12th pass