Question

a projectile is fired with velocity U at an angle theta with horizontal at the highest point of its trajectory it splits up into three segments of masses M and 2m the first part Falls vertically downward with zero initial velocity and second part returns why are the same path to point of projection the velocity of the third pass mass 2m just after the explosion will be

answer with steps​

Answer 1

Answer:

ANSWER

Given, velocity=v,Angle=θ,masses=m,m,2m

So, As the highest point, the velocity of the projectile is ucosθ

Now, the projectile has mass 4m so the momentum possessed by the projectile at the highest point =4mucosθ  (Right Direction)

It breaks into three fragments, one of the fragment falls down so its horizontal momentum is zero.

The second fragment of the mass m return to the same path so it must have the velocity of ucosθ in the opposite direction (left direction)

So,

The fragments of the third fragment of mass 2m, Let the velocity of the third fragment is v,

Thus the momentum of the third fragment is 2mv

Conserving momentum in horizontal dirction,

We have,

pi=pf⇒4mucosθ=−mucosθ+2mv⇒V=25ucosθ

Answer 2

Answer:

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