Problem 52 - Master Resource Book in JEE Main Physics

Question

A particle is projected at \(60^{\circ}\) to the horizontal with an energy \(E .\) The kinetic energy and potential energy at the highest point are [KCET, AIEEE 2007] (a) \(\left(\frac{E}{2}, \frac{E}{2}\right)\) (b) \(\left(\frac{3 E}{4}, \frac{E}{4}\right)\) (c) \((E, 0)\) (d) \(\left(\frac{E}{4}, \frac{3 E}{4}\right)\)

Step-by-Step Solution

Answer

The energies at the highest point are (a) \( (\frac{E}{2}, \frac{E}{2}) \).

Step 1: Understand the Problem

The particle is projected with an initial kinetic energy \(E\) at a \(60^{\circ}\) angle to the horizontal. At its highest point, we need to determine the kinetic and potential energies. We know that at the highest point, the vertical component of velocity is zero, and all kinetic energy is due to the horizontal component.

Step 2: Identify Initial Components of Velocity

The initial velocity can be decomposed into horizontal and vertical components. Let the initial speed be \(v\). The horizontal component is \(v \cos 60^{\circ} = \frac{v}{2}\) and the vertical component is \(v \sin 60^{\circ} = \frac{\sqrt{3}v}{2}\).

Step 3: Determine Energy at Highest Point

At the highest point, the vertical velocity is zero. The kinetic energy is due solely to the horizontal component \(v \cos 60^{\circ}\). Thus, the kinetic energy at the top is \(\frac{1}{2}mv^2 \cos^2 60^{\circ} = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4}mv^2\).

Step 4: Calculate Initial Kinetic Energy

The initial kinetic energy is given as \(E\), which is \(\frac{1}{2}mv^2\). Therefore, \(mv^2 = 2E\).

Step 5: Calculate Kinetic Energy at the Highest Point

From Step 3 and Step 4, the kinetic energy at the highest point is \(\frac{1}{4}mv^2 = \frac{1}{4}(2E) = \frac{E}{2}\).

Step 6: Calculate Potential Energy at the Highest Point

Using energy conservation, the potential energy at the highest point is the initial total energy minus the kinetic energy at that point. Hence, potential energy = \(E - \frac{E}{2} = \frac{E}{2}\).

Step 7: Final Evaluation

The kinetic and potential energies at the highest point are both \(\frac{E}{2}\). Thus, the correct option is (a) \(\left(\frac{E}{2}, \frac{E}{2}\right)\).

Key Concepts

Kinetic Energy

In physics, kinetic energy represents the energy a particle possesses due to its motion. The formula to calculate kinetic energy is \( KE = \frac{1}{2} mv^2 \) , where \( m \) denotes mass and \( v \) is the velocity of the particle. This energy depends on both the mass and speed, showing how much energy an object has when it moves.
Consider a particle projected with an initial kinetic energy \( E \) at an angle to the horizontal. Its initial velocity is divided into horizontal and vertical components. At the top of its trajectory, only the horizontal component contributes to the kinetic energy.

• The horizontal component of the velocity remains constant throughout the projectile's flight.
• Vertical component becomes zero at the highest point of the trajectory.

Thus, the kinetic energy at the highest point is derived solely from the horizontal motion, calculated using \( \frac{1}{4} mv^2 \) , implying half the initial kinetic energy, owing to the squared terms in the wave

Potential Energy

Potential energy is the stored energy of position possessed by an object. For a particle in projectile motion, the potential energy is related to its height above the ground. We calculate it using the formula \( PE = mgh \) , where \( g \) stands for the acceleration due to gravity, and \( h \) is the height.
When the particle reaches its highest point, it has gained the most height. At this peak, the potential energy captures the initial energy that is not used for horizontal kinetic energy. For our projectile, given the total energy initially is \( E \) and has undergone horizontal kinetic energy, its potential energy at the apex is \( E - \frac{E}{2} = \frac{E}{2} \) . This demonstrates energy transformation from kinetic to potential as the projectile rises.

• Potential energy rises as height increases.
• At maximum height, the potential energy is greatest in projectile motion.

Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In projectile motion, the total mechanical energy (kinetic plus potential) remains constant, provided there is no air resistance.
Initially, when the particle is launched, it has maximum kinetic energy and minimum potential energy. As it rises, the kinetic energy decreases while potential energy increases, maintaining the total energy \( E \).

• At launch, high kinetic energy, low potential energy.
• At highest point, energy is equally divided between kinetic and potential forms.
• During descent, potential energy converts back to kinetic.

By understanding energy conservation, we see how the total \( E \) stays the same, providing insights into energy's behavior in projectile motion. It clearly illustrates the interchange between kinetic and potential energies throughout the trajectory.