Problem 10 - Applied Mathematics

Question

State the law of conservation of linear momentum for two interacting particles. Show how the law of conservation of linear momentum applied to two particles which collide directly follows from Newton's laws of motion. Three smooth spheres A, B, C. equal in all respects, lie at rest and separated from one another on a smooth horizontal table in the order \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) with their centres in a straight line. Sphere A is projected with speed \(V\) directly towards sphere \(\mathrm{B}\). If the coefficient of restitution at each collision is \(e\), where \(0

Step-by-Step Solution

Answer

Sphere A transfers velocity, thus leading to indirect collision with C. Sphere A and sphere B will collide second time as A pursues B due to resultant velocities after collision steps.

Step 1: Understand the Law of Conservation of Linear Momentum

The law of conservation of linear momentum states that if no external forces act on a system of interacting particles, the total linear momentum of the system remains constant. Mathematically, for two particles with masses \(m_1\) and \(m_2\) and velocities \(u_1, u_2\) before collision and velocities \(v_1, v_2\) after collision, we have \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Step 2: Applying Newton's Laws of Motion

Newton's Third Law states that for every action, there is an equal and opposite reaction. During the collision, the forces on the two particles are equal in magnitude and opposite in direction. Newton's Second Law states that the change in momentum is equal to the applied force times the duration of the force. When combined, these laws lead to the conclusion that the total momentum of the system is conserved in the absence of external forces.

Step 3: Initial Situation and Definitions

Three smooth spheres A, B, C are placed in a straight line. Sphere A is projected towards sphere B with an initial velocity of \(V\). The masses of the spheres are all equal, say \(m\), and the coefficient of restitution for each collision is \(e\). Initially, the velocities are: \( u_A = V, u_B = 0, u_C = 0 \).

Step 4: First Collision (A and B)

Using the coefficient of restitution \(e\), the relative velocities before and after the collision are related by \[ v_B - v_A = -e (u_A - u_B) \]. Also apply conservation of momentum: \[ m u_A + m u_B = m v_A + m v_B \]. Solve these equations to find the velocities \(v_A\) and \(v_B\) after the first collision.

Step 5: Equation After First Collision

Since \(u_B = 0\), the equations from Step 4 become: \[ v_B - v_A = -e V \], and \[ m V = m v_A + m v_B \]. Solve to get: \[ v_A = 0.5(V - Ve) = V(1-e)/2 \], and \[ v_B = 0.5(V + Ve) = V(1+e)/2 \].

Step 6: Second Collision (B and C)

The velocities before this collision are: \( u_B = V(1+e)/2, u_C = 0 \). Applying similar principles as in Step 4: \[ v_C - v_B' = -e (u_B - u_C) \], conserve momentum: \[ m u_B + m u_C = m v_B' + m v_C \]. After solving these, get: \[ v_B' = V(1+e)e/2 = Ve(1+e)/2 \], and \[ v_C = V(1+e)/2 \].

Step 7: Determining Second Impact of A and B

Noting that \(v_A = V(1-e)/2\) and \(v_B' = Ve(1+e)/2 \), and considering directions, sphere A will surely strike sphere B again because \(v_B' > v_A\). On equating their positions over time, second impact can be inferred.

Step 8: Conclusion

Sphere A strikes sphere B initially, transferring velocity component \(V(1+e)/2\) to B. It then advances to a point where A and B collide again due to \(v_B' being greater \) indicating pursuit and subsequent collision.

Key Concepts

Newton's laws of motion

Newton's laws of motion can establish a foundation to understand many physical interactions. There are three laws:

First, Newton's First Law explains that an object will remain at rest or move in a straight line with constant speed unless acted upon by a force. This reinforces the idea of inertia.

Second, Newton's Second Law is about the relationship between force, mass, and acceleration: \( F = ma \). This tells us that the force applied to an object is equal to the mass of the object multiplied by its acceleration.

Third, Newton's Third Law states that for every action, there is an equal and opposite reaction. This means if object A exerts a force on object B, then object B will exert an equal and opposite force on object A.

In the given problem, when sphere A hits sphere B, sphere B exerts an equal and opposite force back on sphere A. This mutual interaction is where Newton’s Third Law plays a role and the second law relates to how these forces affect the movement (velocities) of the spheres.

coefficient of restitution

The coefficient of restitution (denoted as \( e \)) is a measure of how 'bouncy' a collision is. It is a value between 0 and 1 that describes the relative velocity after and before an impact.

When two objects collide, they may not bounce back completely. The parameter \( e \) is defined as:

\( e = \frac{{\text{{relative velocity after collision}}}}{{\text{{relative velocity before collision}}}} \).

If \( e = 1 \), we have a perfectly elastic collision where no kinetic energy is lost. If \( e = 0 \), we have a perfectly inelastic collision, where the objects stick together.

In the exercise, every collision has a coefficient of restitution \( e \), indicating that the spheres don’t stick together but don’t bounce back perfectly either. The relative velocities after and before collision are adjusted by this factor, playing a key role in determining the final velocities of the spheres.

elastic collisions

Elastic collisions are collisions where there is no loss of kinetic energy in the system. In other words, the total kinetic energy before and after the collision remains the same.

In the context of our problem, although we consider the coefficient of restitution \( e \), showing the collision isn’t perfectly elastic, the concept of elastic collisions helps us understand momentum conservation.

Generally, during an elastic collision:

• Both momentum and kinetic energy are conserved.
• The objects will bounce off each other with no permanent deformation or heat generation.

Since kinetic energy and momentum are conserved, mathematical expressions for these conservation laws are used to find the velocities post-collision. Even with \( e < 1 \), understanding elastic collisions helps in applying momentum conservation laws properly.

momentum conservation in collisions

The conservation of linear momentum is a key principle in collision problems. It states that if no external forces act on a system of particles, the total linear momentum of the system remains constant.

Mathematically, for two colliding particles, it is given by:

\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)

Where:

• \( m_1 \) and \( m_2 \) are the masses of the particles,
• \( u_1 \) and \( u_2 \) are their velocities before collision,
• and \( v_1 \) and \( v_2 \) are their velocities after collision.

For our problem, when sphere A hits sphere B, the total momentum before and after collision is conserved. This principle is used along with the coefficient of restitution to find the velocities after each collision.

The problem then requires solving these equations step by step to find the movement of all three spheres until sphere A hits sphere B again. Using conservation laws helps us track these movements accurately.