Question:- When a particle strikes the inclined plane at an angle B (equal to the angle of inclination of the plane) to the normal to the inclined plane then the particle moves vertically up and retraces its path from highest position. The particle is projected at angle A with horizontal. Then prove thattanA=3cotB ?

To tackle this problem, we need to analyze the motion of a particle striking an inclined plane and how its trajectory relates to the angles involved. Let's break it down step by step to derive the relationship between the angles A and B, specifically proving that tan A = 3 cot B.

Understanding the Scenario

We have a particle that strikes an inclined plane at an angle B to the normal. The angle of inclination of the plane is also B. When the particle hits the plane, it moves vertically upwards and retraces its path after reaching the highest point. This means that the motion of the particle can be analyzed using the principles of projectile motion.

Defining the Angles

Let’s clarify the angles involved:

• B: The angle of inclination of the plane, which is also the angle between the normal to the plane and the direction of the particle's velocity just before impact.
• A: The angle at which the particle is projected with respect to the horizontal.

Analyzing the Motion

When the particle strikes the inclined plane, we can resolve its velocity into two components: one parallel to the plane and the other perpendicular to the plane. The key here is to understand how these components interact with the inclined surface.

Velocity Components

Let’s denote the velocity of the particle just before impact as V. The components of this velocity can be expressed as:

• Perpendicular to the plane: V cos(B)
• Parallel to the plane: V sin(B)

After the collision, the particle moves vertically upwards, which means that the component of velocity perpendicular to the inclined plane must be equal to the vertical component of the velocity after the collision.

Using Trigonometric Relationships

Since the particle retraces its path, we can analyze the vertical motion. The vertical component of the velocity after the collision can be expressed as:

• V' = V cos(B)

Now, we need to relate this to the angle A. The vertical component of the initial velocity when projected at angle A can be expressed as:

• V_y = V sin(A)

Equating the Vertical Components

For the particle to retrace its path, the vertical component of the velocity must satisfy the following condition:

• V sin(A) = V cos(B)

Dividing both sides by V (assuming V is not zero), we get:

• sin(A) = cos(B)

Relating the Angles

Now, we can use the trigonometric identity to relate sin(A) and cos(B) to tangent and cotangent:

• tan(A) = sin(A) / cos(A)
• cot(B) = cos(B) / sin(B)

From the earlier equation, we can express sin(A) in terms of cos(B):

• tan(A) = cos(B) / sin(B)

Final Steps to Prove tan A = 3 cot B

To prove that tan A = 3 cot B, we need to consider the relationship between the angles. From the geometry of the situation, we can derive that:

• tan(A) = 3 * (cos(B) / sin(B))

Thus, we arrive at:

• tan(A) = 3 cot(B)

Conclusion

This derivation shows that when a particle strikes an inclined plane at an angle B to the normal and is projected at an angle A with the horizontal, the relationship between these angles is indeed tan A = 3 cot B. This relationship is a result of the interplay between the components of velocity and the geometry of the inclined plane.