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Chapter 3: Problem 246

Determine the coefficient of restitution \(e\) for a steel ball dropped from rest at a height \(h\) above a heavy horizontal steel plate if the height of the second rebound is \(h_{2}\).

Short Answer

Expert verified
The coefficient of restitution \( e \) is \( \sqrt{\frac{h_{2}}{h}} \).

Step by step solution

01

Understand the Concept

The coefficient of restitution (e) measures how much kinetic energy of a ball is retained after a collision. It relates the velocity of separation to the velocity of approach between two colliding bodies. For a ball dropped from a height h, its rebound height can be used to find e.
02

Determine the velocity just before impact

When the ball falls, its velocity just before impact can be determined using \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity.
03

Determine the velocity just after rebound

The velocity just after rebound can be calculated using \( v' = \sqrt{2gh_{2}} \), where \( h_{2} \) is the rebound height.
04

Find the coefficient of restitution

The coefficient of restitution \( e \) is calculated via the formula:\[ e = \frac{\text{velocity after rebound}}{\text{velocity before impact}} = \frac{v'}{v} \]. Substituting the expressions from Step 2 and Step 3 gives:\[ e = \frac{\sqrt{2gh_{2}}}{\sqrt{2gh}} = \sqrt{\frac{h_{2}}{h}} \].

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Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy associated with the motion of an object. For an object with mass, if it's moving, it means it has kinetic energy. The mathematical formula is simple, given by \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object. This equation tells us that kinetic energy depends quadratically on velocity. This means that doubling the speed of an object will quadruple its kinetic energy!
  • Conservation of Kinetic Energy: In elastic collisions, kinetic energy is conserved. However, in inelastic collisions, some kinetic energy is transformed into other forms of energy like heat or sound.
  • Relation with Coefficient of Restitution: The coefficient of restitution indicates how much kinetic energy is preserved post-collision. It ranges from 0 (perfectly inelastic collision) to 1 (perfectly elastic collision).
In the context of the exercise, when a ball rebounds, not all of the initial kinetic energy is retained, which is quantified by the coefficient of restitution.
Velocity of Separation
The velocity of separation is a concept associated with the relative motion of two bodies post-collision. Essentially, it describes how quickly bodies are moving apart after an impact.
  • This concept is crucial in determining the coefficient of restitution, as it involves calculating how fast objects separate compared to how fast they were coming together.
  • It is given by: \[ V_{sep} = v_2' - v_1' \] where \( v_2' \) and \( v_1' \) are the velocities of the bodies after collision.
In our exercise, considering the steel ball and the heavy plate, the plate remains stationary after collision, so the velocity of separation equates to the rebounding velocity of the ball, which can be calculated from the height of rebound.
Velocity of Approach
The velocity of approach is the rate at which two bodies are closing in on each other before they collide. Just like the velocity of separation, it is vital for figuring out the coefficient of restitution. Knowing how quickly two objects move towards each other gives us insight into the intensity of their forthcoming collision.
  • It is calculated as: \[ V_{app} = v_1 - v_2 \] where \( v_1 \) and \( v_2 \) are the initial velocities of the colliding objects.
  • For objects that are not initially moving but are dropped from a height, as in our steel ball exercise, this velocity can be determined by gravitational equations: \( v = \sqrt{2gh} \), where \( h \) is the height from which the object is dropped.
In our scenario, because the plate doesn't move, the velocity of approach is simply the pre-collision velocity of the falling ball.
Gravitational Acceleration
Gravitational acceleration is a key factor in understanding the motion of bodies falling under the influence of gravity. On Earth, this constant is denoted by \( g \) and is approximately 9.81 m/s虏. It affects the velocity of an object as it falls and impacts its kinetic energy.
  • Importance in Free Fall: In the absence of air resistance, all objects fall with this constant acceleration, making it a crucial element of calculations related to free fall.
  • Effect on Velocity: When an object falls, its velocity just before impact can be determined using \( v = \sqrt{2gh} \). Similarly, post-rebound velocity sheds light on how high the object returns based on gravitational forces.
In our exercise, gravitational acceleration is used to compute the velocity of both approach and separation, making it an essential component for resolving the problem.

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Most popular questions from this chapter

The cart of mass \(m_{1}=3 \mathrm{kg}\) is moving to the right with a speed \(v_{1}=6 \mathrm{m} / \mathrm{s}\) when it collides with the initially stationary barrier of mass \(m_{2}=5\) kg. The coefficient of restitution for this collision is \(e=\) \(0.75 .\) Determine the maximum deflection \(\delta\) of the barrier, which is connected to three springs, each of which has a modulus of \(4 \mathrm{kN} / \mathrm{m}\) and is undeformed before the impact.

The two mine cars of equal mass are connected by a rope which is initially slack. Car \(A\) is given a shove which imparts to it a velocity of \(4 \mathrm{ft} / \mathrm{sec}\) with \(\operatorname{car} B\) initially at rest. When the slack is taken up, the rope suffers a tension impact which imparts a velocity to car \(B\) and reduces the velocity of car \(A\). (a) If 40 percent of the kinetic energy of car \(A\) is lost during the rope impact, calculate the velocity \(v_{B}\) imparted to car \(B\) (b) Following the initial impact, car \(B\) overtakes car \(A\) and the two are coupled together. Calculate their final common velocity \(v_{C}\).

The 0.1 -lb projectile \(A\) is subjected to a drag force of magnitude \(k v^{2},\) where the constant \(k=\) \(0.0002 \mathrm{lb}-\mathrm{sec}^{2} / \mathrm{ft}^{2} .\) This drag force always opposes the velocity \(\mathbf{v} .\) At the instant depicted, \(v=100 \mathrm{ft} / \mathrm{sec}\) \(\theta=45^{\circ},\) and \(r=400 \mathrm{ft} .\) Determine the corresponding values of \(\ddot{r}\) and \(\ddot{\theta}\).

A rocket launches an unpowered space capsule at point \(A\) with an absolute velocity \(v_{A}=8000 \mathrm{mi} / \mathrm{hr}\) at an altitude of 25 mi. After the capsule has traveled a distance of 250 mi measured along its absolute space trajectory, its velocity at \(B\) is \(7600 \mathrm{mi} / \mathrm{hr}\) and its altitude is \(50 \mathrm{mi}\). Determine the average resistance \(P\) to motion in the rarified atmosphere. The earth weight of the capsule is \(48 \mathrm{lb}\), and the mean radius of the earth is 3959 mi. Consider the center of the earth fixed in space.

A 75 -g projectile traveling at \(600 \mathrm{m} / \mathrm{s}\) strikes and becomes embedded in the 40 -kg block, which is initially stationary. Compute the energy lost during the impact. Express your answer as an absolute value \(|\Delta E|\) and as a percentage \(n\) of the original system energy \(E\).
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