Problem 8 - Applied Mathematics

Question

A sphere \(A\), of mass \(m_{1}\), and velocity \(u\), collides with a stationary sphere B of mass \(m_{2}\). If sphere \(\mathrm{A}\) is brought to rest by the collision. find the velocity of \(\mathrm{B}\) after impact. and the coefficient of restitution. If sphere B now collides with a stationary sphere \(\mathrm{C}\) and is brought to rest find the mass of sphere \(\mathrm{C}\) assuming the same coefficient of restitution between \(\mathrm{A}\) and \(\mathrm{B}\), and \(\mathrm{B}\) and \(\mathrm{C}\).

Step-by-Step Solution

Answer

The velocity of sphere B is \( \frac{m_{1}u}{m_{2}} \). The coefficient of restitution is \( \frac{m_{1}}{m_{2}} \). The mass of sphere C is \(m_{1}\).

Step 1 - Understand the Problem

We need to find the velocity of sphere B after sphere A collides with it and comes to rest. Next, we need to find the coefficient of restitution between spheres A and B. Finally, using the coefficient of restitution, we need to find the mass of sphere C, given that sphere B collides with sphere C and comes to rest.

Step 2 - Apply Conservation of Momentum for Sphere A and B Collision

According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision. The initial momentum of sphere A is \(m_{1}u\) and sphere B is stationary, so its initial momentum is 0. After the collision, since sphere A is brought to rest, its final momentum is 0, and the final momentum of sphere B is \(m_{2}v_{B}\). Therefore, the equation is \(m_{1}u = m_{2}v_{B}\).

Step 3 - Solve for Velocity of Sphere B

From the conservation of momentum equation \(m_{1}u = m_{2}v_{B}\), solve for the velocity \(v_{B}\). The velocity of sphere B after impact is \(v_{B} = \frac{m_{1}u}{m_{2}}\).

Step 4 - Apply Coefficient of Restitution (e) Formula

The coefficient of restitution (e) is defined as the relative speed of separation divided by the relative speed of approach. The relative speed of approach before collision is \(u\) (since sphere B is stationary) and the relative speed of separation after collision is \(v_{B}\). So, \(e = \frac{v_{B}}{u} = \frac{\frac{m_{1}u}{m_{2}}}{u} = \frac{m_{1}}{m_{2}}\).

Step 5 - Apply Conservation of Momentum for Sphere B and C Collision

A similar analysis applies when sphere B collides with sphere C. Sphere B has momentum \(m_{2}v_{B}\) and sphere C is initially stationary. The final momentum of sphere B is 0 (since it comes to rest) and the final momentum of sphere C is \(m_{3}v_{C}\). Therefore, the equation is \(m_{2}v_{B} = m_{3}v_{C}\).

Step 6 - Use Coefficient of Restitution for Sphere B and C Collision

Using the coefficient of restitution between sphere B and sphere C, which is the same as between sphere A and B \(e = \frac{m_{1}}{m_{2}}\), the relative speed of separation is \(v_{C}\) and the relative speed of approach is \(v_{B}\). So, \(e = \frac{v_{C}}{v_{B}} = \frac{m_{1}}{m_{2}}\).

Step 7 - Solve for Velocity of Sphere C

From the restitution equation, \(v_{C} = e \times v_{B} = \frac{m_{1}}{m_{2}} \times \frac{m_{1}u}{m_{2}} = \frac{m_{1}^2u}{m_{2}^2}\).

Step 8 - Solve for Mass of Sphere C

Using the momentum conservation equation for spheres B and C, \(m_{2}v_{B} = m_{3}v_{C}\), substitute \(v_{B} = \frac{m_{1}u}{m_{2}}\) and \(v_{C} = \frac{m_{1}^2u}{m_{2}^2}\) to get \(m_{2} \times \frac{m_{1}u}{m_{2}} = m_{3} \times \frac{m_{1}^2u}{m_{2}^2}\). Solve for \(m_{3}\) to get \(m_{3} = m_{1}\).

Key Concepts

Conservation of Momentum

The principle of conservation of momentum is fundamental in collision mechanics. It states that the total momentum of a closed system remains constant if no external forces act on it. In a collision between two spheres, momentum is transferred from one sphere to the other, but the total momentum of both spheres together before and after the collision remains the same.

Consider two spheres: A (mass = m₁, initial velocity = u) and B (mass = m₂, initially at rest). When Sphere A collides with Sphere B and comes to rest right after the impact, the total momentum before the collision (\text{m}_{1}\text{u}) must equal the total momentum after the collision (\text{m}_{2}\text{v}_{B}).

This can be expressed as:
\[\text{m}_{1}\text{u} + 0 = \text{m}_{2}\text{v}_{B}\]
Which simplifies to:
\[\text{v}_{B} = \frac{\text{m}_{1}\text{u}}{\text{m}_{2}}\]
As a result, you can see that the velocity of Sphere B after the impact depends directly on the mass and initial velocity of Sphere A and the mass of Sphere B. When Sphere B subsequently collides with Sphere C and is brought to rest, the momentum conservation principles can similarly be applied.
Remember that in both analyses, we assuming no external forces interfere. So, once you understand conservation of momentum, you can analyze many collision problems confidently.

Coefficient of Restitution

The coefficient of restitution (e) is a measure of how elastic a collision is between two objects. It is defined as the ratio of the relative speed of separation to the relative speed of approach after and before the collision, respectively.

In mathematical terms, for our spheres A and B:
\[e = \frac{\text{velocity after collision of B} - \text{velocity after collision of A}}{\text{initial velocity of A}- \text{initial velocity of B}}\]

Elastic Collisions

In elastic collisions, not only is momentum conserved, but kinetic energy is also conserved.
Elastic collisions are ideal scenarios where no kinetic energy is lost to factors like heat, sound, or material deformation during the collision. Sphere A and B colliding, as described in the problem, is an elastic collision. Here’s a simpler analogy: imagine two perfectly hard spheres (like steel balls) colliding on a smooth surface.
During the collision, the conservation laws enable us to find different post-collision velocities and physical attributes, like the coefficient of restitution, to analyze them step-by-step. Note, although real-world collisions exhibit some loss here, they can often be approximated well enough to elastic cases in problem-solving scenarios.