Question

A body of mass M is projected at an angle Q to the horizontal the projectile at highest point breaks into two fragments of equal masses one of the fragment retraces its path to a point of projection the velocity of the Other fragment is after explosion is ?

Answer 1

Answer:

Explanation:

A body of mass M is projected at an angle Q to the horizontal

=> Horizontal Velocity = V CosQ

Vertical Velocity = VSinQ

Vertical Velocity become 0 at Highest point

so Velocity at highest point = V CosQ

Momentum = MVCosQ

Breaks into two Parts  of Equal masses = M/2 each

MVCosQ = (M/2)Vf  - (M/2)VCosQ

=> 2MVCosQ = MVf  - MVCosQ

=> Vf =  3VCosQ

Answer 2

Answer:

Explanation:

A body of mass M is projected at an angle Q to the horizontal

=> Horizontal Velocity = V CosQ

Vertical Velocity = VSinQ

Vertical Velocity become 0 at Highest point

so Velocity at highest point = V CosQ

Momentum = MVCosQ

Breaks into two Parts of Equal masses = M/2 each

MVCosQ = (M/2)Vf - (M/2)VCosQ

=> 2MVCosQ = MVf - MVCosQ

=> Vf = 3VCosQ