The Question is from NEET guide , and it was created by the author P.NARASIMHA

Question

A particle of mass m is projected with a velocity v making an angle of ${45^0}$ with the horizontal. The magnitude of angular momentum of projectile about the point of projection when the particle is at its maximum height h is

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Angular momentum of a particle about a point.
	**Related Topics**: Projectile motion, Kinematics, Vector cross product.
	**Prerequisites**: Understanding of angular momentum, projectile motion equations.
## SOLUTION PATHWAY
**Given Information**:
	* Mass of particle: m
	* Initial velocity: v
	* Projection angle: 45 degrees
	* Point of reference: point of projection
	* Position: maximum height h
**Method & Steps**:
	1. **Find the horizontal velocity at max height**: The horizontal velocity remains constant throughout the projectile motion. $v_x = v \cos(45^0) = \frac{v}{\sqrt{2}}$
		Reasoning: Horizontal velocity in projectile motion is constant.
		Formula/Rule: $v_x = v \cos(θ)$
	2. **Find the vertical distance (h) at max height**:  $h = \frac{v^2 \sin^2(45^0)}{2g} = \frac{v^2 (1/2)}{2g} = \frac{v^2}{4g}$
		Reasoning: Using the formula for maximum height in projectile motion.
		Formula/Rule: $h = \frac{u^2\sin^2(θ)}{2g}$
	3. **Calculate the angular momentum (L) at max height**: Angular momentum is given by $\vec{L} = \vec{r} 	imes \vec{p}$, where $\vec{r}$ is the position vector and $\vec{p}$ is the linear momentum. Here, position vector will be (x,h) and linear momentum is m*vx in i cap direction. L = r * p * sin(theta), here r is the position vector from the point of projection to the particle at max height. the angle between position vector and the momentum vector is 90 degree. L = r*m*vx. Horizontal distance at max height x = v_x * t_max/2 = v*cos(45)*v*sin(45)/g = v^2/2g. The position vector is r = xi + hj. L = |(x i + hj) x mvxi|. L = m * h * vx. L = m * h * v / sqrt(2).
		Working: L =  m * (v/sqrt(2)) * h = $\frac{mvh}{\sqrt{2}}$
		Key Point: Only the horizontal component of velocity contributes to angular momentum at max height.
	4. **Final step**: The magnitude of angular momentum is  $\frac{mvh}{\sqrt{2}}$
		Result: $\frac{mvh}{\sqrt{2}}$
		Verification: The units of angular momentum are kg m²/s, which is consistent with mvh.
## OPTION ANALYSIS
[A] :x: Incorrect. Angular momentum is not zero at maximum height as the horizontal component of velocity is non-zero.
[B] ✓: Correct. The calculation matches the correct answer. $\frac{mvh}{\sqrt{2}}$
[C] :x: Incorrect. The correct expression does not involve $h^2$.
[D] :x: Incorrect. Option B is correct.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrect use of the position vector.  It must be the vector from the point of projection to the particle at its maximum height.
		2. Forgetting to use the horizontal velocity at max height when calculating linear momentum.
	**Quick Tips**:
		1. Always remember that angular momentum is a vector quantity and the cross product needs to be considered.
		2. The horizontal component of velocity is constant throughout projectile motion.

P.NARASIMHA · Answer

Zero

P.NARASIMHA · Answer

The correct Option is "$\frac{{mvh}}{{\sqrt 2 }}$"

P.NARASIMHA · Answer

$\frac{{mv{h^2}}}{{\sqrt 2 }}$

P.NARASIMHA · Answer

None of these

NEET Questions

Select Subject

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