Final answer:
Correct option: (B). 3/2u cos θ.
To conserve horizontal momentum, the third fragment of mass 2m must have a velocity of 3/2u cos θ immediately after the projectile splits at its peak, thus the correct answer is (B).
Step-by-step explanation:
The student is asking about a special case of conservation of momentum in a two-dimensional projectile motion scenario.
The key information to understand here is that at the peak of its trajectory, a projectile has a horizontal velocity component νux = u cos θ, but its vertical velocity component is zero.
As the projectile splits into three parts, conservation of momentum dictates that the total momentum of the system in each direction (horizontal and vertical) must be conserved.
In the case presented, one segment falls vertically, implying it has no horizontal motion and therefore doesn't affect the horizontal momentum balance.
The second segment returns back to the point of projection, which means it has an initial horizontal velocity equal and opposite to the projectile's initial horizontal velocity.
Hence, for the third part of mass 2m to conserve the total horizontal momentum, it must have a velocity 3u cos θ to balance the momentum taken by the second part and maintain the horizontal momentum contributed by both the first (zero) and second parts.
Therefore, the correct answer to the student's question is (B). 3/2u cos θ.