Problem 47 - Master Resource Book in JEE Main Physics

Question

A body is projected with speed \(v \mathrm{~ms}^{-1}\) at angle \(\theta\). The kinetic energy at the highest point is half of the initial kinetic energy. The value of \(\theta\) is (a) \(30^{\circ}\) (b) \(45^{\circ}\) (c) \(60^{\circ}\) (d) \(90^{\circ}\)

Step-by-Step Solution

Answer

The angle \( \theta \) is \( 45^{\circ} \) (option b).

Step 1: Understand the Problem

We need to determine the angle \( \theta \) of projection such that the kinetic energy at the highest point is half of the initial kinetic energy of the body.

Step 2: Initial Kinetic Energy Calculation

The initial kinetic energy, \( KE_i \), can be calculated using the formula \( KE_i = \frac{1}{2} m v^2 \), where \( m \) is the mass of the body and \( v \) is the initial speed. However, mass \( m \) will ultimately cancel out, so we focus on \( v \).

Step 3: Kinetic Energy at Highest Point

At the highest point of projection, the vertical component of velocity is zero. The horizontal component of velocity, \( v_x = v \cos \theta \), remains. Thus, the kinetic energy at the highest point, \( KE_h \), is \( KE_h = \frac{1}{2} m (v \cos \theta)^2 = \frac{1}{2} m v^2 \cos^2 \theta \).

Step 4: Set up the Kinetic Energy Ratio Equation

According to the problem, the kinetic energy at the highest point is half of the initial kinetic energy: \( \frac{1}{2} m v^2 \cos^2 \theta = \frac{1}{2} \times \frac{1}{2} m v^2 \).

Step 5: Simplify the Equation

Cancel the common terms and simplify the equation: \( \cos^2 \theta = \frac{1}{2} \).

Step 6: Solve for Angle \( \theta \)

The equation \( \cos^2 \theta = \frac{1}{2} \) implies \( \cos \theta = \frac{1}{\sqrt{2}} \). Therefore, \( \theta = 45^{\circ} \).

Step 7: Choose the Correct Option

The calculated angle \( \theta \) is \( 45^{\circ} \), which corresponds to option (b).

Key Concepts

Kinetic Energy

Kinetic energy is a crucial concept in physics, especially when studying projectile motion. It refers to the energy an object possesses due to its motion. In simple terms, the faster or heavier an object is, the more kinetic energy it has. The formula to calculate kinetic energy is given by \[ KE = \frac{1}{2} m v^2 \]where

• \( KE \) is the kinetic energy,
• \( m \) is the mass of the object, and
• \( v \) is its velocity.

In projectile motion, kinetic energy can change as the velocity of the object changes. Initially, an object has kinetic energy due to its initial speed. As the object moves along its path, these energy levels can vary. For instance, at the highest point of a projectile's path, the vertical velocity component becomes zero. The horizontal velocity component, however, remains constant. This enables us to determine the kinetic energy at the peak.

Angles of Projection

When studying projectile motion, the angle of projection is vital. It determines the trajectory, range, and behavior of the projectile during flight. The angle of projection, often denoted as \( \theta \), is the angle between the initial velocity vector and the horizontal axis.Different angles lead to different motion characteristics:

• Low Angles: Produce longer range but shorter maximum height.
• Medium Angles (around \( 45^\circ \)): Achieve the maximum range due to the ideal balance between height and distance.
• High Angles (approaching \( 90^\circ \)): Lead to high vertical motion, but limited range.

For example, in our exercise, the question involves finding an angle where the kinetic energy at a particular point (the highest point) is precisely half of the initial kinetic energy. Solving the equation for \( \theta \) gives us the critical angle where these conditions hold true.

Horizontal and Vertical Components of Velocity

Understanding the horizontal and vertical components of velocity is essential in solving problems related to projectile motion. When a projectile is launched, its velocity can be split into two perpendicular components:

• Horizontal Component (\( v_x \)): Given by \( v \cos \theta \), it remains constant throughout the motion because there is no acceleration acting in the horizontal direction (assuming air resistance is negligible).
• Vertical Component (\( v_y \)): Given by \( v \sin \theta \), it changes over time due to gravitational acceleration. At the highest point, \( v_y \) becomes zero.

These components help to analyze different stages of projectile motion independently. In our exercise, the kinetic energy calculation at the highest point requires the understanding of these components. At this peak, the kinetic energy depends solely on the horizontal component of the velocity since the vertical component is zero.