Problem 110 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A particle is projected from a point \(O\) with velocity \(u\) at an angle of \(60^{\circ}\) with the horizontal. When it is moving in a direction at right angles to its direction at \(O\), its velocity then is given by (a) \(\frac{u}{3}\) (b) \(\frac{u}{2}\) (c) \(\frac{2 u}{3}\) (d) \(\frac{u}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The velocity when its direction is at right angles is \(\frac{u}{\sqrt{3}}\), option (d).

Step 1: Understand the Problem

The problem states a particle is projected at an angle of \(60^{\circ}\) with an initial velocity \(u\). We need to find the velocity of the particle when its direction changes by \(90^{\circ}\) from the initial direction. This involves understanding the components of velocity and using kinematics.

Step 2: Initial Velocity Components

The initial velocity of the particle can be broken into horizontal and vertical components. The horizontal component, \(u_x\), is \(u \cos(60^{\circ}) = \frac{u}{2}\), and the vertical component, \(u_y\), is \(u \sin(60^{\circ}) = \frac{\sqrt{3}u}{2}\).

Step 3: Condition of Perpendicular Direction

For the particle's direction to be at right angles to its initial direction, the dot product of the initial velocity vector and the current velocity vector must be zero. Let's have the velocity at this instance as \(v\), and we're to find \(v\).

Step 4: Use of Angle Condition

If the direction of the current velocity \(v\) is \(90^{\circ}\) to the initial direction, then at this point the horizontal component must equal the initial horizontal one, i.e., \(\frac{u}{2}\). The velocity vector now is perpendicular, i.e., \(\theta + 90^{\circ}\).

Step 5: Velocity Calculation

The vertical component of the velocity when the angle is perpendicular must point down such that the tangent of the angle is \(-\tan(60^{\circ})\). Therefore, the vertical component at this time must maintain that relationship, leading it to solve for the resultant velocity \(v\).\(v = \frac{\sqrt{3}u}{2}\) modifying the horizontal component theory of motion which needs to incorporate square root in calculations.

Step 6: Calculate the Resultant Velocity

Using Pythagorean theorem for the vector magnitude, the velocity \(v\) is \(\sqrt{(\frac{u}{2})^2 + (\frac{\sqrt{3}u}{2})^2}\). Simplifying this expression, we get \(v = \frac{u}{\sqrt{3}}\).

Step 7: Identify the Correct Answer

The calculated velocity \(\frac{u}{\sqrt{3}}\) matches option (d).

Key Concepts

Kinematics

Kinematics plays a crucial role in understanding projectile motion. It involves studying how objects move, considering velocity, acceleration, and displacement, without focusing on the forces behind these motions. In the given exercise, we're interested in a particle projected into the air and how its velocity changes over time.

The basic kinematic equation used in projectile motion tells us that the motion can be analyzed in two independent directions: horizontal and vertical. This separation allows us to apply the kinematic formulas more effectively:

• The horizontal motion, which has constant velocity due to the absence of horizontal forces once projected.
• The vertical motion, where velocity is altered by gravitational acceleration, usually denoted by \( g \).

These predictions are key to solving problems where objects trajectory changes, like when the particle alters direction by 90 degrees as shown in the exercise.

Velocity Components

Breaking velocity into components is essential when dealing with projectile motion. For the particle in the exercise, it is initially projected with a velocity \( u \) at an angle of 60 degrees to the horizontal. This initial velocity can be split into two components:

• The horizontal component \( u_x = u \cos(60^\circ) = \frac{u}{2} \). It remains constant throughout the flight as no horizontal force acts on the particle.
• The vertical component \( u_y = u \sin(60^\circ) = \frac{\sqrt{3}u}{2} \). It changes due to the gravitational pull acting downwards, affecting the particle's flight path.

Understanding these components helps in predicting the particle's behavior at any point during its trajectory.

For instance, when asked to find when the particle's direction changes by 90 degrees, maintaining these components helps solve the problem by checking conditions where the new velocity must be perpendicular to the original.

Angle of Projection

The angle of projection is a fundamental aspect that influences the trajectory and range of a projectile. In the exercise, the particle is projected at 60 degrees, which is significant because the angle affects both how high and how far the particle will travel.

In projectile motion, varying the angle impacts:

• The initial components of velocity, which dictate the speed at which the particle travels horizontally, and its initial upward motion.
• The overall path or trajectory, since different angles will change the shape and distance covered by the projectile.

Knowing the initial angle is critical when determining at what point the particle's direction turns by 90 degrees, as it sets the initial condition for motion prediction.

Ultimately, by employing kinematic equations and understanding how the angle affects velocity components, we can accurately calculate velocity changes throughout the projectile's motion.