Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle \(60 \degree\)with horizontal, it can travel a distance \(x_1\) along the plane. But when the inclination is decreased to \(30 \degree\) and the same object is shot with the same velocity, it can travel \(x_2\) distance. Then \(x_1:x_2\) will be:

• A. \(1:\sqrt2\)
• B. \(\sqrt2:1\)
• C. \(1:\sqrt3\)
• D. \(1:2\sqrt3\)

Answer 1

The Correct Option is C

To solve this problem, we will analyze the motion of the object on an inclined plane using the components of the velocity and gravitational force. Let's proceed step by step:

Step 1: Understand Problem Setup

When an object is projected up an inclined plane, the maximum distance it can travel along the plane depends on the angle of inclination (\(\theta\)) and the initial velocity (\(v_0\)). The forces acting on the object include the gravitational force component along the plane.

Step 2: Derive Formula for Maximum Distance

If an object is projected with an initial velocity \(v_0\) up an incline at an angle \(\theta\), the distance (\(x\)) it can travel along the plane until it comes to a stop can be derived using the kinematic equation:

v^2 = u^2 - 2a s

where:

• \(v = 0\) (since the object comes to a stop)
• \(u = v_0\)
• \(a = g \sin \theta\) (acceleration due to gravity component along the incline)
• \(s = x\) (distance along the plane)

The equation becomes:

0 = v_0^2 - 2g \sin \theta \cdot x

Solving for \(x\), we have:

x = \frac{v_0^2}{2g \sin \theta}

Step 3: Calculate Distance Ratios

1. For \(\theta = 60^\circ\): x_1 = \frac{v_0^2}{2g \sin 60^\circ} = \frac{v_0^2}{2g \left(\frac{\sqrt{3}}{2}\right)} = \frac{v_0^2}{g\sqrt{3}}
2. For \(\theta = 30^\circ\): x_2 = \frac{v_0^2}{2g \sin 30^\circ} = \frac{v_0^2}{2g \left(\frac{1}{2}\right)} = \frac{v_0^2}{g}

Step 4: Determine Ratio \(x_1 : x_2\)

Now, find the ratio:

\frac{x_1}{x_2} = \frac{\left(\frac{v_0^2}{g\sqrt{3}}\right)}{\left(\frac{v_0^2}{g}\right)} = \frac{1}{\sqrt{3}}

Thus, the ratio \(x_1 : x_2 = 1 : \sqrt{3}\).

Conclusion

The correct option is therefore \(1:\sqrt{3}\). This is because, for the same initial velocity, the maximum range on a less steep plane is longer due to the reduced effect of gravitational force along the plane.