A projectile is fired with velocity u at an angle @ with horizontal.At the hi

e total mass of the particle = 4m

initial velocity= u

angle with horizontal=@

vertical component of velocity=usin(@)

horizontal component of velocity=ucos(@)

let velocity at the highest point= v

and the velocity of fragment of mass 2m =w

at the highest point,there is no vertical velocity

i.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 zero
hence momentum in horizontal direction is conserved
i.e P(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 2mvcos

ghst 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 of mass 2m just after explosion will be what???

nidhi

8 Years agoGrade 12th pass