Problem 50 - Master Resource Book in JEE Main Physics

Question

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I A point particle of mass \(m\) moving with speed \(v\) collides with stationary point particle of mass \(M\). If the maximum energy loss possible is given as \(f\left(\frac{1}{2} m v^{2}\right)\), then \(f=\left(\frac{m}{M+m}\right)\) Statement II Maximum energy loss occurs when the particles get stuck together as a result of the collision. [UEE Main 2013] (a) Statement 1 is true, Statement \(\|\) is true, and Statement II is the correct explanation of Statement 1 (b) Statement 1 is true, Statement \(\|\) is true, but Statement II is not the correct explanation of Statement 1 (c) Statement \(I\) is true, Statement \(\|\) is false (d) Statement 1 is false, Statement Il is true

Step-by-Step Solution

Answer

Statement I is false, Statement II is true. (d) is correct.

Step 1: Understanding energy loss in a collision

For a collision, the maximum energy loss occurs when the two particles stick together and move as one body after the collision. The kinetic energy before the collision is solely due to the moving mass, and after the collision, it's shared between the combined mass.

Step 2: Expressing initial and final kinetic energy

Before the collision, the total kinetic energy,\[ KE_{initial} = \frac{1}{2} m v^2. \]After the collision, the two masses move together with a velocity \( V \). By conservation of momentum, \( mv = (M+m)V \).

Step 3: Finding the velocity after collision

By rearranging the momentum equation:\[ V = \frac{mv}{M+m}. \]

Step 4: Calculate final kinetic energy

Using the combined velocity:\[ KE_{final} = \frac{1}{2} (M+m) \left( \frac{mv}{M+m} \right)^2 = \frac{1}{2} \frac{m^2 v^2}{M+m}. \]

Step 5: Calculate energy loss

The energy loss is given by the difference between the initial and final kinetic energy:\[ \text{Energy Loss} = \frac{1}{2} m v^2 - \frac{1}{2} \frac{m^2 v^2}{M+m}. \]Simplifying this expression gives:\[ \text{Energy Loss} = \frac{1}{2} m v^2 \left[ 1 - \frac{m}{M+m} \right] = \frac{1}{2} m v^2 \frac{M}{M+m}. \]

Step 6: Verification of Statement I

Statement I claims that the maximum energy loss possible is \( f\left(\frac{1}{2} m v^{2}\right) \) where \( f=\frac{m}{M+m} \). From our derivation, the maximum energy loss factor is actually \( \frac{M}{M+m} \), not \( \frac{m}{M+m} \). Thus, Statement I is false.

Step 7: Verification of Statement II

Statement II claims that maximum energy loss occurs when the particles get stuck together. This is true, as seen from our derivation where sticking together leads to maximum energy dissipation.

Key Concepts

Energy Conservation

In physics, energy conservation refers to the idea that the total energy in a closed system remains constant over time. During a collision, the focus is often on kinetic energy, which is the energy of motion. Before a collision, each object has its own kinetic energy, and after the collision, kinetic energy is shared among the objects involved. This sharing depends on the nature of the collision.

In an elastic collision, kinetic energy is conserved, and the total kinetic energy before and after the collision remains the same. However, in inelastic collisions, some of the kinetic energy is transformed into other forms of energy, like heat or sound. Thus, not all kinetic energy is retained as motion after the collision. The key point in energy conservation during collisions is understanding that the sum of energies, including converted forms, remains constant.

The principle helps in understanding the dynamics of the system and is crucial for calculating unknown variables, such as velocities after the collision.

Momentum Conservation

Momentum conservation is a fundamental principle that states the total momentum of a closed system is constant if the system is not influenced by external forces. In the context of collisions, momentum conservation is an invaluable tool for predicting and analyzing the outcome of the collision.

Momentum is the product of mass and velocity, expressed by the formula: \[ p = mv \]
In an isolated system where two objects collide, like in our exercise, the total momentum before the collision is equal to the total momentum after the collision. This can be articulated mathematically for our scenario as:\[ mv = (M+m)V \]where:

• \( m \) is the mass of the moving particle
• \( M \) is the mass of the stationary particle
• \( v \) is the initial velocity of the moving particle
• \( V \) is the common velocity after they stick together

Understanding this concept is crucial as it allows us to determine post-collision velocities, showing that momentum conservation is key to analyzing collision outcomes.

Inelastic Collision

Inelastic collisions are a type of collision where the colliding objects do not separate after impact. Instead, they move together as one mass post-collision, which is why they are often called completely or perfectly inelastic collisions.

In such a collision, the maximum amount of kinetic energy is lost through the process of sticking together. The energy lost manifests in other forms, for example, transformed into internal energy, sound, or other non-mechanical energies.

A key feature of inelastic collisions is that while kinetic energy is not conserved, momentum is. Hence, the final motion of the bodies can be calculated using the principles of momentum conservation. The problem we reviewed is such an inelastic collision scenario, where the total kinetic energy is not conserved but decreases by a predictable factor depending on the masses involved.

Kinetic Energy Loss

Kinetic energy loss during a collision refers to the amount of energy that is no longer observed as kinetic movement post-collision. Especially in inelastic collisions, this loss is significant as the bodies stick together and move with a shared, usually reduced, velocity.

The original exercise showcases a formula to quantify this loss:\[ \text{Energy Loss} = \frac{1}{2} m v^2 - \frac{1}{2} \frac{m^2 v^2}{M+m} \]which simplifies to:\[ \text{Energy Loss} = \frac{1}{2} m v^2 \frac{M}{M+m} \]allowing us to see that kinetic energy loss depends on both the initial kinetic energy and the mass ratio of the involved particles.

This energy reduction accounts for energy now observable in non-kinetic forms like heat. Understanding energy loss is pivotal in many applications, from car crash analyses to particle physics experiments, as it governs how energy transforms in everyday and extreme interactions.