Problem 7 - Applied Mathematics

Question

A sphere \(\mathrm{A}\), of mass \(2 m\) and velocity \(2 u\), overtakes and collides with sphere \(\mathrm{B}\), of mass \(m\) and velocity \(u\) travelling in the same line which is perpendicular to a vertical smooth wall. After being struck by A. sphere B goes on to strike the wall. If the coefficient of restitution between \(\mathrm{A}\) and \(\mathrm{B}\) is \(\frac{1}{2}\) and that between \(\mathrm{B}\) and the wall is \(\frac{3}{4}\) show that there is a second collision between \(\mathrm{A}\) and \(\mathrm{B}\) and describe what happens after the second impact.

Step-by-Step Solution

Answer

After the first collision, velocities of A and B are \(\frac{3u}{2}\) and \(\frac{5u}{2}\). B's velocity after rebounding off the wall is \(-\frac{15u}{8}\). There is a second collision since B moves towards A.

Step 1: Define Initial Conditions

Given that sphere A has mass 2m and velocity 2u, and sphere B has mass m and velocity u. The coefficient of restitution (e) between A and B is 1/2.

Step 2: Use Conservation of Momentum for First Collision

Law of conservation of momentum before and after collision: \[ 2m(2u) + m(u) = 2m(v_A) + m(v_B) \] Solving for velocities after the collision, where \( v_A \) and \( v_B \) are the velocities of A and B after the collision.

Step 3: Apply Coefficient of Restitution Formula for First Collision

The coefficient of restitution (e) is given by: \[ e = \frac{\text{relative velocity after collision}}{\text{relative velocity before collision}} = \frac{v_B - v_A}{2u - u} \] Substituting the given value, e = 1/2: \[ \frac{v_B - v_A}{u} = \frac{1}{2} \] Thus, \[ v_B - v_A = \frac{u}{2} \]

Step 4: Solve for Velocities After the First Collision

Using the equations from Step 2 and Step 3, solve for \( v_A \) and \( v_B \). We have two equations: \[ 4mu + mu = 2mv_A + mv_B \] and \[ v_B - v_A = \frac{u}{2} \] Solving these equations together gives us: \[ v_A = \frac{3u}{2} \] and \[ v_B = \frac{5u}{2} \]

Step 5: Calculate Velocity of Sphere B After Impact with Wall

Use the coefficient of restitution between B and the wall, which is 3/4: \[ e = \frac{\text{relative velocity after impact}}{\text{relative velocity before impact}} = \frac{v'_B - 0}{v_B - 0} \] Given that the wall is stationary: \[ v'_B = -e \times v_B = -\frac{3}{4} \times \frac{5u}{2} = -\frac{15u}{8} \]

Step 6: Determine if Second Collision Occurs

Sphere A moves with velocity \( \frac{3u}{2} \) and sphere B after rebounding from the wall has velocity \( -\frac{15u}{8} \). Since \( -\frac{15u}{8} < \frac{3u}{2} \), this indicates that B moves towards A. Therefore, a second collision will occur.

Step 7: Velocities After Second Collision

Using the coefficient of restitution for the second collision (still 1/2 between A and B): \[ e = \frac{v'_B - v'_A}{\left| -\frac{15u}{8} - \frac{3u}{2} \right|} \] Solving: \[ \frac{v'_B - v'_A}{\frac{21u}{8}} = \frac{1}{2} \] Also apply conservation of momentum as before:

Key Concepts

conservation of momentum

Conservation of Momentum is a fundamental concept in physics, especially in collision dynamics. It states that the total momentum of a closed system remains constant before and after a collision, provided no external forces act on it. This principle is derived from Newton's laws of motion.
In our exercise, we analyze the collision between two spheres, A and B. Sphere A has mass 2m and velocity 2u, whereas sphere B has mass m and velocity u. Using the conservation of momentum formula, we can set up the following initial equation:

\[ 2m(2u) + m(u) = 2m(v_A) + m(v_B) \]
Here, \( v_A \) and \( v_B \) are the velocities of A and B after the first collision. By solving for these velocities, we can determine how both spheres behave post-impact. The law ensures that the total momentum before collision (\(2m(2u) + m(u)\)) is equal to the total momentum after collision (\(2m(v_A) + m(v_B)\)). Using this principle simplifies the process of solving collision dynamics problems.

coefficient of restitution

Coefficient of Restitution (\(e\)) measures the elasticity of a collision between two bodies. It is defined as the ratio of the relative velocity after collision to the relative velocity before collision. Mathematically, it is represented as:
\[ e = \frac{v_f - u_f}{u_i - v_i} \]
In the given exercise, the coefficient of restitution between spheres A and B is \(\frac{1}{2}\). This means the relative speed after the collision is half of that before the collision. The formula we use here is:
\[ e = \frac{v_B - v_A}{2u - u} = \frac{1}{2} \]
This simplifies to:
\[ v_B - v_A = \frac{u}{2} \]
This information is crucial as it helps us determine the velocities of A and B after their first collision. The coefficient of restitution provides insights into the nature of the collision—whether it is perfectly elastic (e=1), perfectly inelastic (e=0), or somewhere in between.

impulse and impact in physics

Impulse and Impact in Physics are related concepts often discussed in collision dynamics. Impulse is the product of force and the time over which it acts, changing an object’s momentum. It is given by:
\[ \text{Impulse} = F \times t \]
For collisions, the impulse experienced by the objects is the change in their momentum. Impact refers to the actual event of collision where forces are applied.
In our exercise, when sphere A hits sphere B, an immediate force acts over a short time, resulting in a change in momentum. The impulse experienced by both spheres is equal and opposite due to Newton’s third law of motion. Understanding impulse helps bridge the gap between force and momentum, explaining how velocities of A and B transform due to collision.

kinematics

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces causing them. It involves parameters like displacement, velocity, and acceleration.
In our collision problem, we use kinematics to understand how sphere A with velocity 2u and sphere B with velocity u interact. By examining their velocities before and after the collision, we apply the concepts of linear motion to predict the outcome.
For instance, after the first collision, we derived:
\[ v_A = \frac{3u}{2} \]
\[ v_B = \frac{5u}{2} \]
These results from kinematics help us predict the motion paths of the spheres as they move and rebound. By understanding kinematics, you can decompose complex collision problems into manageable components and solve them step-by-step.

mechanics

Mechanics is the broader field encompassing both kinematics and dynamics to analyze the motion and apply forces to objects. It splits into two main areas: statics (studying objects at rest) and dynamics (studying objects in motion).
In this exercise, we primarily deal with dynamics. Mechanics allows us to apply fundamental principles - like the conservation of momentum and the coefficient of restitution - to solve for velocities and predict collisions.
Moreover, understanding rebound impacts and second collisions becomes clearer through mechanics. As sphere B rebounds from the wall and moves back towards sphere A, we again apply dynamic principles. The second collision's velocities can be calculated using:
\( e = \frac{v'_B - v'_A}{\text{initial relative velocity}} \)
Mechanics combines all these concepts to provide a comprehensive understanding of the collision dynamics.