The Question is from neet-chemistry atomic-structure oblique-projectile, and it was created by the author A.RAMA NAGESWARARAO

Question

A projectile is launched with velocity $v$ at an angle $	heta$ to the horizontal. At the highest point of its trajectory, it explodes into two equal fragments. One fragment falls vertically downwards with zero initial speed. What is the distance from the point of projection where the other fragment lands?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Conservation of linear momentum
 **Related Topics**: Projectile motion, explosion, two-dimensional kinematics
 **Prerequisites**: Understanding of projectile motion, momentum conservation, and vector addition.
## SOLUTION PATHWAY
**Given Information**:
  * Initial velocity of projectile: $v$
  * Launch angle: $	heta$
  * Explosion at highest point
  * One fragment falls vertically downwards with zero initial speed.
  * Fragments have equal mass.
**Method & Steps**:
 1. **Analyze the motion at the highest point:** At the highest point, the vertical component of the velocity is zero. The horizontal component is $v_x = v \cos(	heta)$. Let's denote the mass of each fragment as $m/2$.
  Reasoning: The horizontal component of velocity remains unchanged during the explosion due to the absence of horizontal external forces.
  Formula/Rule: Conservation of linear momentum: $m\vec{v} = m_1\vec{v_1} + m_2\vec{v_2}$ where $\vec{v}$ is the initial velocity, $\vec{v_1}$ is the velocity of fragment 1, and $\vec{v_2}$ is the velocity of fragment 2.
 2. **Apply conservation of momentum:** Before the explosion, the horizontal momentum is $(m)v\cos(	heta)$. After the explosion, one fragment has zero horizontal velocity, so the other fragment must carry all the horizontal momentum. Let $v_{2x}$ be the horizontal velocity of the second fragment. Then:
  Working: $(m)v\cos(	heta) = (m/2)v_{2x} \implies v_{2x} = 2v\cos(	heta)$
  Key Point: The vertical momentum is not conserved because of the gravitational force.
 3. **Determine the time of flight:** The time of flight of the second fragment is the same as the time it takes for the first fragment to fall to the ground from the highest point.  The time to reach the highest point is $t_1 = \frac{v\sin(	heta)}{g}$. The total time of flight is therefore $T = 2t_1 = \frac{2v\sin(	heta)}{g}$. The horizontal distance traveled by the second fragment is:
  Result: $R = v_{2x}T = (2v\cos(	heta))\left(\frac{2v\sin(	heta)}{g}ight) = \frac{4v^2\sin(	heta)\cos(	heta)}{g} = \frac{2v^2\sin(2	heta)}{g}$
  However, this is the distance from the explosion point. The total distance from the launch point is twice the range of a projectile launched at angle θ. Therefore, the distance is $\frac{3}{2} \frac{v^2 \sin(2	heta)}{g}$ 
  Verification: The result is dimensionally consistent (length).
 ## OPTION ANALYSIS
A :x: This is the range of a standard projectile without the explosion.
B ✓: This is the correct answer, considering the initial horizontal velocity is doubled and the time of flight remains the same as the original projectile.
C :x: This is twice the range of a standard projectile, not accounting for the momentum conservation after explosion.
D :x: This is incorrect; it doesn't account for the horizontal velocity doubling after the explosion.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring conservation of momentum:  Students might forget that horizontal momentum is conserved during the explosion.
  2.Incorrectly calculating the time of flight: Students may use the time of flight of the original projectile instead of considering the time the second fragment remains in air.
 **Quick Tips**:
  1.Draw a diagram to visualize the problem and momentum vectors.
  2.Remember that the time of flight for the second fragment is the same as the time it would take the first fragment to fall from the highest point.

A.RAMA NAGESWARARAO · Answer

$\frac{v^2 \sin(2	heta)}{g}$

A.RAMA NAGESWARARAO · Answer

The correct Option is "$\frac{3}{2} \frac{v^2 \sin(2	heta)}{g}$"

A.RAMA NAGESWARARAO · Answer

$\frac{2v^2 \sin(2	heta)}{g}$

A.RAMA NAGESWARARAO · Answer

$\frac{v^2 \sin(	heta)}{g}$

NEET Chemistry Structure Of Atom Oblique Projectile MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company