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Chapter 8: Problem 47

Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).

Short Answer

Expert verified
(a) 0 m/s, (b) 3 m/s each in opposite directions, (c) 2 m/s and 4 m/s.

Step by step solution

01

Understand the Collision Types

There are three different collision scenarios provided: (a) inelastic collision where the two objects stick together, (b) perfectly elastic collision, and (c) an inelastic collision with a coefficient of restitution of \( \frac{1}{3} \). We will analyze each scenario separately.
02

Calculate the Speed After Inelastic Collision

(a) Because the balls stick together after the collision, they move as a single unit. The formula for the velocity of the combined mass is derived from momentum conservation: \[ m u + (-m u) = (m + m) v \]Since both masses are the same and they have equal and opposite velocities, the total momentum before collision is 0. Thus:\[ 0 = 2m v \]\[ v = 0 \]This means the speed of both together after impact is 0.
03

Analyze the Perfectly Elastic Collision

(b) When the collision is perfectly elastic, both momentum and kinetic energy are conserved. From conservation of momentum:\[ m u + (-m u) = m v_1 + m v_2 \]From conservation of kinetic energy:\[ \frac{1}{2} m u^2 + \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]Solving these equations, specifically due to symmetry and equal mass, leads to the velocities post-collision being equal in magnitude but opposite in direction. So, \[ v_1 = -u \] and \[ v_2 = u \],Resulting speeds are \(3 \text{ m/s}\) for both, just in opposite directions from their incoming path.
04

Evaluate the Collision with Restitution Coefficient \(\frac{1}{3}\)

(c) The coefficient of restitution \(e = \frac{1}{3}\) relates initial and final relative velocities:\[ e = \frac{v_2 - v_1}{u_1 - u_2} \], where for head-on collision \(u_1 - u_2 = 6 \text{ m/s}\). Applying this gives:\[ \frac{1}{3} = \frac{v_2 - v_1}{6} \]\[ v_2 - v_1 = 2 \]Using the momentum equation again:\[ 6 = v_1 + v_2 \]Solving these equations, we find:\[ v_1 = 2 \text{ m/s}, \quad v_2 = 4 \text{ m/s}\]. Both move in the initial direction of their respective incoming paths.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inelastic Collision
Inelastic collisions are those in which the colliding objects stick together after the impact. This means they move as a single combined mass post-collision. During such an event, the total momentum of the system is conserved, but the kinetic energy is not. This loss of kinetic energy occurs because it is often transformed into other energy forms like heat or sound.
For example, when two equal mass balls collide and stick together, their velocities cancel each other out, resulting in a net velocity of zero. Here, despite the initial motion, the total kinetic energy is lower post-collision compared to pre-collision.
Perfectly Elastic Collision
A perfectly elastic collision is characterized by the conservation of both momentum and kinetic energy. In such collisions, no kinetic energy is lost to other forms of energy during the impact.
For instance, if two balls collide elastically, their speeds will remain the same, but their directions may change. Imagine two identical balls moving towards one another with the same speed. After a perfectly elastic collision, they will rebound with the same speed but in opposite directions. This mode of collision is ideal and represents a scenario where all kinetic energy is preserved.
Coefficient of Restitution
The coefficient of restitution is a measure that describes how elastic a collision is, with values ranging between 0 and 1. A value of 1 corresponds to a perfectly elastic collision, whereas 0 indicates a perfectly inelastic collision where objects stick together.
It is calculated as the ratio of relative velocity after the collision to the relative velocity before the collision. In a head-on collision, this can be expressed as:
  • \[e = \frac{v_2 - v_1}{u_1 - u_2}\]
For example, if the coefficient is \(\frac{1}{3}\), it suggests that the collision is partially elastic, with some kinetic energy conserved, but not all.
Momentum Conservation
Momentum conservation is a fundamental principle in physics, stating that in a closed system free of external forces, the total momentum before the collision equals the total momentum after.
The formula for momentum is given by the product of mass and velocity, or \(p = mv\). In a collision context, this principle allows us to calculate post-collision velocities by ensuring the sum of momenta before and after remains constant.
  • For the example of two colliding balls with equal mass, their individual momenta perfectly cancel out if they collide head-on with equal speed but opposite direction.
Kinetic Energy Conservation
In some collisions, most typically perfectly elastic ones, kinetic energy is conserved. This implies that the total kinetic energy of the system before collision is equal to the total kinetic energy after.
Kinetic energy can be represented with the formula:
  • \[KE = \frac{1}{2} mv^2\]
In a perfectly elastic collision between two objects, both the velocities and directions could change post-collision, but the system's kinetic energy remains constant. However, this isn't the case for inelastic collisions, where some energy is converted into other forms.

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Most popular questions from this chapter

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