Problem 37 - Master Resource Book in JEE Main Physics

Question

A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of \(60^{\circ}\) with the vertical with the same initial speed. At highest point of their journey, the ratio of their potential energies will be (a) \(1: 1\) (b) \(2: 1\) (c) \(3: 2\) (d) \(4: 1\)

Step-by-Step Solution

Answer

(d) 4:1

Step 1: Understanding Potential Energy at the Highest Point

The Potential Energy (PE) of an object at the highest point is given by the formula \( PE = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the maximum height. Our goal is to find the ratio of the potential energies of the two balls at their highest points.

Step 2: Maximum Height for Vertical Throw

For the ball thrown vertically, the potential energy at the maximum height depends solely on the initial speed \( u \). Use the energy conservation principle: at the highest point, all initial kinetic energy \( \left(\frac{1}{2} mu^2\right) \) converts into potential energy. Thus, \( mgh_1 = \frac{1}{2}mu^2 \), resulting in \( h_1 = \frac{u^2}{2g} \).

Step 3: Maximum Height for Projectile at Angle

For the second ball, thrown at an angle \(60^{\circ}\) with the vertical, decompose the initial speed into vertical component \( u_y = u \cos 60^{\circ} \). The maximum height for the angled throw is calculated using this component: \( h_2 = \frac{u_y^2}{2g} = \frac{(u \cdot \cos 60^{\circ})^2}{2g} = \frac{(u \cdot \frac{1}{2})^2}{2g} = \frac{u^2}{8g} \).

Step 4: Calculate Ratio of Potential Energies

With the heights calculated, ratio of potential energies \( \frac{PE_1}{PE_2} = \frac{mgh_1}{mgh_2} = \frac{h_1}{h_2} = \frac{\frac{u^2}{2g}}{\frac{u^2}{8g}} = \frac{4}{1} \). Hence, the potential energy ratio is \(4:1\).

Key Concepts

Potential Energy

Potential energy is a type of stored energy that is dependent on the position of an object within a field. In the context of a ball thrown upwards, it relates to the height the ball achieves against the force of gravity.

To calculate potential energy (PE) at a particular height, use the formula:

• \( PE = mgh \)

Where:

• \( m \) is the mass of the ball,
• \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth),
• \( h \) is the height above the ground.

Potential energy increases with height. Therefore, higher the ball goes, more potential energy it possesses at its highest point.

In projectile motion, potential energy at a point provides insight into how much kinetic energy has been converted during the ascent.

Vertical Motion

Vertical motion is a type of motion wherein an object moves in a straight line along the vertical axis. When we discuss a ball projected straight up, it travels only in the vertical direction before reversing back down.

Initially, the ball has kinetic energy provided by its initial speed until gravity acts upon it, slowing it down until it stops momentarily at the highest point, where potential energy is maximized.
Some key points on vertical motion include:

• The object's initial speed significantly impacts how high it will travel.
• After reaching maximum height, the object will fall back down, accelerating due to gravity.

Understanding vertical motion is crucial for calculating how high an object can go given its initial speed, which influences potential energy computations.

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, this principle plays an essential role.

In the case of an object projected upwards:

• Kinetic energy, coming from an object's initial speed, is transformed to potential energy as the object rises.
• At the highest point of its trajectory, the object's kinetic energy is nearly all converted to potential energy, which means the object momentarily stops moving (speed = 0 m/s).

As the object falls back down, the potential energy will convert back into kinetic energy. Energy conservation allows us to predict variables like height if given initial speed, since total energy remains constant.

Maximum Height Calculation

Calculating the maximum height an object achieves in projectile motion allows us to determine the potential energy at this peak point. For a vertical throw, the entire initial kinetic energy converts to potential energy at the highest point.
For a ball thrown vertically, this is calculated by the formula:

• \( h_1 = \frac{u^2}{2g} \)

where \( u \) is the initial speed and \( g \) is gravity.
For an object thrown at an angle (e.g., 60 degrees with respect to vertical), we only consider the vertical component of velocity when it comes to height. This is calculated as:

• \( h_2 = \frac{u_y^2}{2g} = \frac{(u \cdot \cos \theta)^2}{2g} \)

Sometimes, a projectile launched at an angle will achieve a lower maximum height due to velocity's division into both vertical and horizontal components. Accurate heights are essential for determining potential energy ratios at peak points, and predict different journey outcomes depending on launch angles.