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A projectile is fired with velocity u at an angle theta with horizontal. At the highest point of its

trajectory it splits into three 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

a) U cos theta b) 3/2 u cos theta c) 2u cos theta d) 5/2 u cos theta

Please find below the solution to the asked query:

At 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 3 fragments.

One of the fragment falls down so its horizontal momentum is zero

the second fragment of mass m returns to the same path, hence it must have the velocity of ucosθ in the opposite direction that is the left direction.

so the momentum of the 2nd fragment will be -mucosθ

let the velocity of the third mass 2m be v in right direction

thus the momentum of the third particle will be 2mv

**Conserving momentum in the horizontal direction**:

we have

**P**

_{i}=P_{f}4mucosθ = -mucosθ +2mv

or 5mucosθ = 2mv

so v = (5/2)ucosθ in the right direction.

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