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Question #42076

Difficulty: 8/10 | 🟢 Free

The Question

A projectile of mass $3M$ is fired from the origin on level ground with an initial speed $u = 50 \text{ m/s}$ at an angle $\theta = \sin^{-1}(4/5)$ above the horizontal. At the highest point of its trajectory, an internal explosion breaks it into two fragments of masses $M$ and $2M$ . Immediately after the explosion, the lighter fragment $M$ moves such that it lands exactly at the point of projection. Assuming the ground is horizontal and $g = 10 \text{ m/s}^2$ , at what distance from the point of projection does the heavier fragment land?

Options

A.
$360 \text{ m}$
B.
$240 \text{ m}$
C.
$300 \text{ m}$
D.
$480 \text{ m}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Projectile Path Equation|System vs Particle Displacement|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the principle of independence of center of mass motion from internal forces.

A common trap is attempting to solve for the individual velocities of the fragments using momentum conservation and then applying kinematic equations.

This often leads to arithmetic errors or forgetting to add the initial horizontal displacement of $120 \text{ m}$ to the heavier fragment's subsequent travel (yielding the trap answer of $240 \text{ m}$ ).

By recognizing that the internal explosion does not alter the CoM trajectory, students can bypass complex kinematics and directly use the CoM coordinate formula at the instant of landing.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Projectile Path Equation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free