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Question

A particle is projected with velocity $v_0$ at an angle $	heta$ with the horizontal. At the highest point of its trajectory, it explodes into two fragments of masses $m_1$ and $m_2$. The lighter fragment $m_1$ falls vertically downwards with zero initial speed.  If the range of the original projectile was $R$, the distance from the point of projection where the heavier fragment lands is:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of linear momentum at the highest point of projectile motion.
	**Related Topics**: Projectile motion, fragmentation, center of mass.
	**Prerequisites**: Understanding momentum conservation, projectile kinematics, and range formula.
## SOLUTION PATHWAY
**Given Information**:
		- Initial velocity: $v_0$ at angle $	heta$.
		- At the highest point, the projectile splits into $m_1$ (falls vertically) and $m_2$.
		- Original range: $R = \frac{v_0^2 \sin 2	heta}{g}$.
**Method & Steps**:
	1. **Horizontal velocity at explosion**:
		Reasoning: At the highest point, vertical velocity is zero. Horizontal velocity remains $v_0 \cos 	heta$.
		Formula: $v_{	ext{horizontal}} = v_0 \cos 	heta$.
	2. **Conservation of momentum**:
		Working: Total momentum before explosion = Total momentum after explosion.
		Key Point: Lighter fragment ($m_1$) has zero horizontal velocity after explosion.
		Equation: $(m_1 + m_2)v_0 \cos 	heta = m_2 v_2$.
	3. **Solve for $v_2$**:
		Result: $v_2 = \left(1 + \frac{m_1}{m_2}ight)v_0 \cos 	heta$.
	4. **Time to fall**:
		Time from highest point to ground: $t = \frac{v_0 \sin 	heta}{g}$ (same as ascent time).
	5. **Total horizontal distance for $m_2$**:
		Distance from explosion: $v_2 \cdot t = \left(1 + \frac{m_1}{m_2}ight)v_0 \cos 	heta \cdot \frac{v_0 \sin 	heta}{g}$.
		Simplification: This equals $\left(1 + \frac{m_1}{m_2}ight)\frac{R}{2}$ (since $\frac{R}{2} = \frac{v_0^2 \sin 	heta \cos 	heta}{g}$).
	6. **Total distance**:
		Result: $\frac{R}{2} + \left(1 + \frac{m_1}{m_2}ight)\frac{R}{2} = R\left(1 + \frac{m_1}{2m_2}ight)$.
## OPTION ANALYSIS
[A] ✓: Correct. Derived directly from conservation of momentum and projectile motion calculations.
[B] :x: Incorrect. Overestimates the momentum ratio by a factor of 2.
[C] :x: Incorrect. Flips the mass ratio ($m_2/m_1$ instead of $m_1/m_2$).
[D] :x: Incorrect. Represents only half the original range, ignoring the horizontal motion of $m_2$ after explosion.
## LEARNING AIDS
	**Common Mistakes**:
		1. Forgetting to conserve momentum horizontally at the highest point.
		2. Incorrectly calculating the time of flight after the explosion.
	**Quick Tips**:
		1. At the highest point, only horizontal velocity matters for momentum conservation.
		2. Use $R = \frac{v_0^2 \sin 2	heta}{g}$ to relate the range to velocity components.

VENKATESWARARAO · Answer

The correct Option is "R(1 + m1/(2*m2))"

VENKATESWARARAO · Answer

R(1 + 2*m1/m2)

VENKATESWARARAO · Answer

R(1 + m2/(2*m1))

VENKATESWARARAO · Answer

R/2

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