Question

A particle is projected from bottom of inclined plane with speed u find distance covered along plane before coming to rest :

• A. \(\frac{u^2}{2g \sin\theta}\)
• B. \(\frac{u^2}{g \sin\theta}\)
• C. \(\frac{u^2}{g \cos\theta}\)
• D. \(\frac{u^2}{2g \cos\theta}\)

Hint:

For problems on inclined planes, it is almost always easiest to resolve forces and motion into components parallel and perpendicular to the plane.
The acceleration of any object sliding on a smooth inclined plane (due to gravity alone) is always \(g \sin\theta\) directed down the plane. Memorizing this can save time.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A particle is given an initial velocity \(u\) up a smooth inclined plane. We need to find the distance it travels along the incline before its velocity becomes zero.
Step 2: Key Formula or Approach:
We will use one of the equations of motion for constant acceleration: \(v^2 = u^2 + 2as\). To use this, we first need to determine the acceleration (\(a\)) of the particle along the inclined plane.
Step 3: Detailed Explanation:
First, let's find the acceleration. The only force acting on the particle along the incline (assuming it's smooth, i.e., no friction) is the component of gravity directed down the plane.
The force of gravity is \(mg\), acting vertically downwards. The component of this force parallel to the incline is \(mg \sin\theta\).
This force opposes the motion of the particle, so it's a retarding force. Using Newton's second law, \(F = ma\):
\[ ma = -mg \sin\theta \] \[ a = -g \sin\theta \] The negative sign indicates that the acceleration is directed down the plane, opposite to the initial velocity.
Now, we use the kinematic equation \(v^2 = u^2 + 2as\):
- Initial velocity (along the plane) = \(u\).
- Final velocity (at the highest point) \(v = 0\).
- Acceleration (along the plane) \(a = -g \sin\theta\).
- Distance covered along the plane = \(s\).
Substitute these values into the equation:
\[ (0)^2 = (u)^2 + 2(-g \sin\theta)s \] \[ 0 = u^2 - 2gs \sin\theta \] \[ 2gs \sin\theta = u^2 \] Now, solve for the distance \(s\):
\[ s = \frac{u^2}{2g \sin\theta} \] Step 4: Final Answer:
The distance covered by the particle along the plane before coming to rest is \(\frac{u^2}{2g \sin\theta}\).