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18. A projectile is fired with velocity u at an angle ewith horizontal. At the highest point of its trajectoryit splits up into three segments of masses m, mand 2 m. First part falls vertically downward withzero initial velocity and second part returns viasame path to the point of projection. The velocityof third part of mass 2 m just after explosion will be(1) u cos032) Ucos252(3) 2u cose(4) ucose

Answer»

the 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(


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