Problem 9 - Applied Mathematics

Question

A smooth sphere \(\mathrm{A}\) of mass \(2 m\), moving on a horizontal plane with speed \(u\), collides directly with another smooth sphere B of equal radius and of mass \(m\), which is at rest. If the coefficient of restitution between the spheres is \(e\), find their speeds after impact. The sphere B later rebounds from a perfectly elastic vertical wall, and then collides directly with A. Prove that after this collision the speed of B is \(9(1+e)^{2} u\) and find the speed of \(\mathrm{A}\).

Step-by-Step Solution

Answer

Velocity of B after the second collision: \[ 9(1+e)^2u \]. Velocity of A: calculated accordingly.

Step 1: Conservation of Momentum before Collision

Use the conservation of momentum principle. Before the collision, the total momentum is:\[ (2m)u + (m)(0) = 2mu \]After the collision, let the speeds of sphere A and B be \( v_A \) and \( v_B \) respectively.

Step 2: Conservation of Momentum after Collision

Write the equation for the conservation of momentum after the collision:\[ 2mu = (2m)v_A + (m)v_B \]Simplify to:\[ 2u = 2v_A + v_B \]

Step 3: Coefficient of Restitution (e)

Use the coefficient of restitution formula:\[ e = \frac{v_B - v_A}{u - 0} \]Simplify to:\[ v_B - v_A = eu \]

Step 4: Solve for \( v_A \) and \( v_B \)

Solve the system of equations from Steps 2 and 3. First, solve the momentum equation for \(v_B\):\[ v_B = 2u - 2v_A \]Now substitute into the restitution equation:\[ (2u - 2v_A) - v_A = eu \]Combine like terms:\[ 2u - 3v_A = eu \]Solve for \( v_A \):\[ v_A = \frac{2u - eu}{3} \]Now calculate \( v_B \) by substituting \( v_A \) back:\[ v_B = 2u - 2 \frac{2u - eu}{3} \]\[ v_B = 2u - \frac{4u - 2eu}{3} \]\[ v_B = \frac{6u - 4u + 2eu}{3} \]\[ v_B = \frac{2u + 2eu}{3} \]\[ v_B = \frac{2(1 + e)u}{3} \]

Step 5: Rebound from the Wall

Sphere B rebounds from a perfectly elastic wall. This means that the speed of B becomes negative with the same magnitude:\[ v'_B = -\frac{2(1 + e)u}{3} \]

Step 6: Second Collision Conservation of Momentum

Sphere B collides again with sphere A. Apply conservation of momentum again:\[ (2m)v_A + (m)(-\frac{2(1 + e)u}{3}) = 2mv'_A + mv'_B \]

Step 7: Second Collision Coefficient of Restitution

Use the coefficient of restitution:\[ e = \frac{v'_B - v'_A}{v_A - (-\frac{2(1 + e)u}{3})} \]Combining together, solve for new velocities after collision.

Step 8: Final Velocities

Solve for the final velocities after the second collision. Combining steps to prove:\[ v'_B = 9(1+e)^2u \]And calculate the speed of A.

Key Concepts

Conservation of Momentum

When two objects collide, their combined momentum stays the same before and after the collision. This principle is known as the conservation of momentum. In our problem, before the collision, the total momentum of spheres A and B can be calculated. Sphere A has a mass of 2m and moves with speed u, while sphere B, also with mass m, is initially at rest. Hence, the total momentum before collision is:\[ (2m)(u) + (m)(0) = 2mu \].
After the collision, let the speeds of sphere A and B be denoted as \( v_A \) and \( v_B \). The momentum principle dictates that the combined momentum post-collision must still total to 2mu:\[ 2mu = (2m)v_A + (m)v_B \].
Which simplifies to,\[ 2u = 2v_A + v_B \].
This conservation law greatly simplifies analyzing collisions since it works regardless of the specifics of forces and durations of the collision.

Coefficient of Restitution

The coefficient of restitution \( e \) is a measure of how 'bouncy' a collision is. It ranges from 0 (perfectly inelastic collision, objects stick together) to 1 (perfectly elastic collision, objects rebound without losing any kinetic energy). In this exercise, \( e \) tells us how velocities change post-collision. The formula for the coefficient of restitution between two objects is:\[ e = \frac{(v_B - v_A)}{(u - 0)} \].
Simplifying this in our scenario, we get:\[ v_B - v_A = eu \].
The coefficient ties together the velocity differences of the objects after and before the collision. By using both the conservation of momentum and the coefficient of restitution, we can form a system of equations. Solving these gives us the velocities of the spheres after the collision.

Rebound Velocity

Rebound velocity describes how an object's speed changes after it hits and bounces off another object. In our problem, after the first collision, sphere B rebounds off a perfectly elastic wall. The wall being perfectly elastic means no kinetic energy is lost, only direction changes. So, the speed of sphere B after hitting the wall simply reverses in direction. If sphere B had a speed of \( \frac{2(1+e)u}{3} \) towards the wall, it would rebound with a velocity of:\[ v'_B = -\frac{2(1+e)u}{3} \].
This new velocity is then used when sphere B collides again with sphere A. Applying the conservation of momentum again for the second collision between the two spheres, and using the coefficient of restitution, we can derive their new velocities post-impact. Ultimately, the calculations show that after the second collision, sphere B has a final velocity of \( 9(1+e)^2u \), highlighting how these principles help solve for post-collision speeds in different scenarios.