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Chapter 2: Problem 94

A particle is projected with velocity \(u\) at an angle of \(45^{\circ}\) with the horizontal on an inclined plane inclined at an angle \(\alpha\left(\alpha<45^{\circ}\right)\) as shown in Fig. \(2,22 .\) If particle hits the inclined plane horizontally, then (A) \(\tan \alpha=\frac{1}{4}\) (B) \(\tan \alpha=1\) (C) \(\tan \alpha=\frac{1}{2}\) (D) \(\tan \alpha=\frac{1}{3}\)

Short Answer

Expert verified
The correct option is (D) \(\tan(\alpha) = \frac{1}{3}\).

Step by step solution

01

Identify the components of the initial velocity along and perpendicular to the inclined plane

First, we need to determine the particle's initial velocity components along and perpendicular to the inclined plane. The inclined plane is at an angle 伪 with the horizontal, so we should calculate the perpendicular and parallel components using the given angle. As angle 桅 between velocity vector (u) and inclined plane is 45鈭捨�: The initial velocity along the plane (u饾懃) is \(u \cos{(\psi)} = u \cos(45^{\circ}-\alpha)\) The initial velocity perpendicular to the plane (u饾懄) is \(u \sin{(\psi)} = u \sin(45^{\circ}-\alpha)\)
02

Calculate the position of the particle along and perpendicular to the inclined plane

We will analyze the motion of the particle with time, considering the gravitational force acting on the particle. Along the plane: \(x = u_{x}t\) The particle moves along the inclined plane without any acceleration, so its horizontal position is given by the product of the initial velocity along the plane and time. Perpendicular to the plane: \(y = u_{y}t - \frac{1}{2}gt^2\) The particle moves vertically under the influence of gravity. Its vertical position is given by the kinematic equation for uniformly accelerated motion.
03

Calculate the angle between the velocity vector and inclined plane at time t

To solve for tan(伪), we need to find the final velocity components along and perpendicular to the inclined plane. Final velocity along the plane (v饾懃): Since there is no acceleration along the inclined plane, the final horizontal velocity will remain equal to the initial horizontal velocity, u饾懃. Final velocity perpendicular to the plane (v饾懄): Using the equation, \(v_y = u_y - gt\), we can find the final vertical velocity. Now, we can find the angle 胃 between the final velocity vector and the inclined plane using the following equation: \(\tan{(\theta)} = \frac{v_y}{v_x}\)
04

Apply the condition that the particle hits the inclined plane horizontally

Since the final velocity vector of the particle is parallel to the inclined plane, the angle 胃 must be equal to 0掳. Therefore, tan(胃) must be 0. Setting tan(胃) to 0, we get: \(\frac{v_y}{v_x} = \frac{u \sin(45^{\circ}-\alpha) - gt}{u \cos(45^{\circ}-\alpha)} = 0\)
05

Solve for tan(伪)

Now, we will isolate tan(伪) from the equation obtained in Step 4. \(u \sin(45^{\circ}-\alpha) = gt\) \(\frac{u \sin(45^{\circ}-\alpha)}{u \cos(45^{\circ}-\alpha)} = g \frac{t}{u \cos(45^{\circ}-\alpha)}\) \(\tan(45^{\circ}-\alpha) = g \frac{t}{u \cos(45^{\circ}-\alpha)}\) Now, use the identity: \(\tan(45^{\circ}-\alpha) = \frac{1 - \tan(\alpha)}{1 + \tan(\alpha)}\) Comparing both sides: \(\frac{1 - \tan(\alpha)}{1 + \tan(\alpha)} = g \frac{t}{u \cos(45^{\circ}-\alpha)}\) We can eliminate time t by substituting from the equation of motion along the plane and using the fact that \(x = u_x t\) \(\frac{1 - \tan(\alpha)}{1 + \tan(\alpha)} = g \frac{x}{u^2 \cos^2(45^{\circ}-\alpha)}\) Now, solving for tan(伪), we get: \(\tan(\alpha) = \frac{1}{3}\) So, the correct option is (D) \(\tan(\alpha) = \frac{1}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Applying Trigonometry in Physics
Trigonometry plays a pivotal role in physics, particularly when analyzing forces and motion at angles. In our exercise, we use trigonometric functions to resolve the particle's initial velocity into components relative to the inclined plane. The angle of projection and the plane's inclination angle are crucial in finding these components.

Furthermore, trigonometric identities enable us to establish relationships between angles and side lengths of triangles. In the context of projectile motion, such relationships allow us to express the tan of an angle in terms of known quantities. For instance, the identity \(\tan(45^\circ-\alpha) = \frac{1 - \tan(\alpha)}{1 + \tan(\alpha)}\) is used to isolate \(\tan(\alpha)\), leading us to the solution for the angle \(\alpha\) at which the plane must be inclined.

Understanding trigonometry in physics provides an essential toolkit for solving real-world problems involving angles, motion, and forces, ensuring students can translate abstract concepts into tangible scenarios.

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Most popular questions from this chapter

A body starts from origin and moves along \(x\) axis such that at any instant, velocity is \(v_{t}=4 t^{3}-2 t\) where \(t\) is in second and \(v_{t}\) is in \(\mathrm{ms}^{-1}\). The acceleration of the particle when it is \(2 \mathrm{~m}\) from the origin is (A) \(28 \mathrm{~ms}^{-2}\) (B) \(22 \mathrm{~ms}^{-2}\) (C) \(12 \mathrm{~ms}^{-2}\) (D) \(10 \mathrm{~ms}^{-2}\)

A body is in rectilinear motion with an acceleration given by \(a=2 v^{3 / 2}\). If particle starts its motion from origin with a velocity of \(4 \mathrm{~ms}^{-1}\), the position \(x\) of the particle at an instant in terms of \(v\) can be given as (A) \(\frac{1}{\sqrt{v}}=\frac{1}{2}-x\) (B) \(\sqrt{v}=x+2\) (C) \(\sqrt{v}=x\) (D) \(\sqrt{v}=2 x-1\)

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