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A shell in flight explodes into n equal ...

A shell in flight explodes into `n` equal fragments `k` of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

A

g

B

(n-k)g

C

`((n-k)g)/(k)`

D

`((n-k)g)/(n)`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the situation of a shell that explodes into `n` equal fragments and understand the behavior of the center of mass of these fragments after the explosion. ### Step-by-Step Solution: 1. **Understanding the Explosion**: - A shell (or bomb) is in flight and explodes into `n` equal fragments. - The explosion causes the fragments to move in different directions, but the total mass of the system remains constant. 2. **Identifying the Fragments**: - Out of the `n` fragments, `k` fragments reach the ground earlier than the remaining `n-k` fragments. - This indicates that the fragments have different trajectories and times of flight after the explosion. 3. **Center of Mass Concept**: - The center of mass (COM) of a system of particles is defined as the point where the total mass of the system can be considered to be concentrated. - The motion of the center of mass is determined by the total external forces acting on the system. 4. **Acceleration of the Center of Mass**: - Regardless of how the fragments move after the explosion, the acceleration of the center of mass is influenced only by external forces. - In this case, the only significant external force acting on the fragments is gravity (assuming air resistance is negligible). 5. **Conclusion**: - Since the only force acting on the system is gravity, the acceleration of the center of mass will be equal to the acceleration due to gravity, denoted as `g`. - Therefore, the acceleration of the center of mass after the explosion remains `g`. ### Final Answer: The acceleration of the center of mass subsequently will be **g**.
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