Problem 20 - Applied Mathematics

Question

An inelastic string has a particle \(\mathrm{A}\) attached to one end and a particle \(\mathrm{B}\) attached to the other end. If \(\mathrm{A}\) is projected in the direction \(\overrightarrow{\mathrm{BA}}\) find the initial speed of \(\mathrm{B}\) if: (a) initially the string is slack. (b) the speed of projection of \(\mathrm{A}\) is \(4 \mathrm{~ms}^{-1}\), (c) the particles are of equal mass, (d) the string is \(2 \mathrm{~m}\) long.

Step-by-Step Solution

Answer

The initial speed of B is 4 m/s.

Step 1: Understand the Problem

Identify the given information and what needs to be found. The string has two particles, A and B, of equal mass. A is projected at a speed of 4 m/s and initially the string is slack. The goal is to find the initial speed of B.

Step 2: Define the Initial Conditions

Since the string is initially slack, particle A starts moving freely until the string is taut. When the string becomes taut, particle B will start to move.

Step 3: Apply the Conservation of Momentum

As the string becomes taut, the momentum in the system must be conserved. Let the mass of particles A and B be m. The conservation of momentum principle states that the total momentum before and after the string becomes taut must be equal.

Step 4: Set Up the Equations

Before the string becomes taut, only particle A is moving. The initial momentum of the system is given by: \[ \text{Initial Momentum} = m \times 4 \text{ m/s} \]After the string becomes taut, both particles A and B will move. Let the speed of B be v. The final momentum of the system is:\[ \text{Final Momentum} = m \times v + m \times (-v) = 0 \]

Step 5: Solve for the Initial Speed of B

Since the momentum has to be conserved and given only particle A is moving before the string becomes taut, we have:\[ m \times 4 \text{ m/s} = m \times v \]Solving for v,\[ v = 4 \text{ m/s} \]

Key Concepts

Inelastic Collisions

Inelastic collisions are a type of collision where the colliding objects stick together after impact. This means that kinetic energy is not conserved in these collisions, even though momentum is. In our exercise, we see this concept in action when particle A, after being projected, eventually makes the string taut, causing both particles to start moving together. When the string tightens, energy is lost due to the deformation, but momentum must still be conserved, as per the laws of physics.

Understanding inelastic collisions helps to explain real-world situations, like car crashes, where two cars might crumple and stick together. The progressive loss of kinetic energy in such scenarios can explain significant phenomena, even if intuitively momentum seems complicated.

Kinematics

Kinematics is the study of motion without considering the forces that cause it. It's a branch of mechanics that defines concepts like velocity, acceleration, displacement, and time. In this exercise, kinematic principles help understand how particle A moves initially at 4 m/s in a straight line towards particle B.

We assume the string is slack, which means particle A will travel unhindered until the string tightens. This scenario is straightforward in terms of motion; linear and uniform, defined by the velocity provided (4 m/s).

• Initial speed of particle A: 4 m/s
• Movement direction: straight line towards B
• No acceleration or deceleration considered

Kinematic equations and these initial conditions deliver the necessary foundation to solve the problem.

Momentum Conservation

Momentum conservation is a fundamental principle in physics stating that the total momentum of a closed system remains constant if no external forces act on it. In our problem, we rely on momentum conservation to solve for the final speeds after the string becomes taut.

Initially, only particle A has momentum, calculated as:

• Initial Momentum: \( m \times 4 \text{ m/s} \)

When the string tightens, both particles move, and their momenta must add up to the initial momentum of particle A. Since the particles have equal mass and they start to move together, this leads to:

\[ m \times 4 \text{ m/s} = m \times v \]

The beautiful simplicity of momentum conservation helps us directly solve for the unknown speed, establishing the uniform applicability of these principles in both simple and complex systems.

Projectile Motion

Projectile motion typically involves analyzing the trajectory of an object thrown or projected into the air, under the influence of gravity. However, in our problem, while the term 'projection' is used, the motion described is a simpler one-dimensional movement. Still, understanding general projectile motion can be very instructive.

The main components of projectile motion analysis include:

• Initial velocity
• Angle of projection
• Time of flight
• Maximum height and range

Here, particle A's movement toward B can be seen as a horizontal projectile, with initial speed but no vertical movement. Simplifying assumptions like these make such problems more straightforward, showcasing the fundamental concepts of motion in a controlled manner.