Problem 45 - Master Resource Book in JEE Main Physics

Question

A jet airplane travelling at a speed of \(500 \mathrm{~km} / \mathrm{h}\) ejects its products of combustion at the speed of \(1500 \mathrm{~km} / \mathrm{h}\) relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? [NCERT] (a) \(-1000 \mathrm{~km} / \mathrm{h}\) (b) \(1000 \mathrm{~km} / \mathrm{h}\) (c) \(100 \mathrm{~km} / \mathrm{h}\) (d) \(-100 \mathrm{~km} / \mathrm{h}\)

Step-by-Step Solution

Answer

The speed of the combustion products with respect to the ground is \(-1000\, \mathrm{km/h}\) (Option a).

Step 1: Understand the Problem

We need to find the speed of the combustion products with respect to an observer on the ground. The given speeds are relative to the jet plane and the ground.

Step 2: Identify Given Information

We have the speed of the jet airplane relative to the ground as \(500\, \mathrm{km/h}\) and the speed of the combustion products relative to the jet airplane as \(1500\, \mathrm{km/h}\).

Step 3: Apply Relative Speed Formula

The speed of the combustion products relative to the ground is calculated by adding the speed of the jet to the relative speed of the combustion products: \(v_{comb-ground} = v_{jet-ground} + v_{comb-jet}\).

Step 4: Plug In Values

Substitute the given speeds into the formula: \(v_{comb-ground} = 500 + 1500 = 2000\, \mathrm{km/h}\).

Step 5: Correct the Equation Interpretation

Note that the combustion products are ejected backwards, thus the relative speed should be negative: \(v_{comb-ground} = 500 - 1500 = -1000\, \mathrm{km/h}\).

Step 6: Final Interpretation

The negative sign implies the combustion products move in the opposite direction to the plane's motion, so the speed relative to the ground is \(-1000\, \mathrm{km/h}\).

Key Concepts

Jet Propulsion

Jet propulsion is a fascinating concept that powers not just airplanes, but also rockets and some modern vehicles. At the core of jet propulsion is Newton's Third Law of Motion: for every action, there is an equal and opposite reaction.
When a jet engine expels gas, it pushes against the air at high speed in one direction. This action causes the jet to move in the opposite direction.

• The high-speed expulsion of combustion gases provides thrust, propelling the jet forward.
• This is why jet propulsion is highly effective for high-speed travel, such as that of military jets and commercial planes.
• The speed of the exhaust relative to the aircraft affects the overall propulsion effect achieved.

In the given problem, the jet expels its combustion products at a speed relative to itself, which influences how observers will perceive the speed of those combustion products based on their frame of reference.

Frame of Reference

Understanding the frame of reference is critical when solving problems involving relative velocity. A frame of reference is simply the perspective from which an observer measures or experiences something.
When we say the jet is traveling at a speed of 500 km/h, this is relative to the ground – the reference frame being used here is ground-based.

• Different frames of reference can vastly change the perceived velocity of an object.
• If you are inside the jet, your frame of reference is different, leading to different velocity perceptions.
• In our problem, the combustion products' speed is provided relative to the plane, not the ground.

So, when calculating the speed of the combustion products as observed from the ground, you have to account for the speed of the plane and the speed of the combustion products as observed from the plane. Understanding how different frames of reference contribute to relative motion is key to solving such problems.

Velocity Addition

Velocity addition is a technique used to determine the relative velocity of an object in different frames of reference. In simpler terms, it helps us find out how fast something is moving relative to a specific reference point.
In physics, the formula to add velocities depends on the frames from which these velocities are measured. The basic formula is:\[ v_{relative} = v_{object} + v_{frame} \]

• This formula tells us how velocities combine when an object moves within a moving reference frame.
• It is essential to consider the direction of motion, as velocities can either add up or cancel out depending on their directions.
• In our problem, the combustion products have a negative velocity relative to the ground because they are ejected in the opposite direction of the jet's movement.

Therefore, the speed of the jet is added negatively to the speed of the combustion products when viewed from the ground, resulting in the velocity of -1000 km/h.