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A \(70-g\) ball \(B\) dropped from a height \(h_{0}=1.5 \mathrm{m}\) reaches a height \(h_{2}=0.25 \mathrm{m}\) after bouncing twice from identical 210 -g plates. Plate \(A\) rests directly on hard ground, while plate \(C\) rests on a foam- rubber mat. Determine \((a)\) the coefficient of restitution between the ball and the plates, \((b)\) the height \(h_{1}\) of the ball's first bounce.

Short Answer

Expert verified
(a) Coefficient of restitution is approximately 0.64. (b) First bounce height is approximately 0.615 m.

Step by step solution

01

Understanding the Problem

Start by understanding the key quantities and what is being asked. The ball is dropped from a height of 1.5 m, and after two bounces on identical plates, it reaches a height of 0.25 m. We need to find the coefficient of restitution between the ball and the plates and the height of the first bounce.
02

Using the Coefficient of Restitution Formula

The coefficient of restitution, denoted by \(e\), is the ratio of the velocity after impact to the velocity before impact relative to the surfaces at impact. For a ball bouncing, \(e = \sqrt{ \frac{h_{f}}{h_{i}} }\), where \(h_{f}\) is the rebound height, and \(h_{i}\) is the height from which it was dropped.
03

Calculating the Effective Coefficient of Restitution

The effective coefficient of restitution for two bounces can be calculated as follows: \[ e_{eff} = \sqrt{ \frac{h_2}{h_0} }. \]Therefore, \[ e_{eff} = \sqrt{ \frac{0.25}{1.5} } \approx \sqrt{0.1667} \approx 0.408. \] This coefficient represents the two sequential bounces.
04

Estimating the Coefficient of Restitution for One Bounce

The effective coefficient of restitution for one bounce can be derived from the effective coefficient for two bounces by solving \[ e^2 = e_{eff}. \]Solving for \(e\), we get:\[ e = \sqrt{e_{eff}} = \sqrt{0.408} \approx 0.64. \] This is the coefficient of restitution for one bounce.
05

Calculating the Height of the First Bounce

To find the height \(h_1\) of the first bounce, use the formula \[ h_1 = e^2 \cdot h_0. \]Substituting the values, \[ h_1 = (0.64)^2 \cdot 1.5 \approx 0.64 \cdot 1.5 = 0.6144 \approx 0.615 \text{ m}. \]

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Key Concepts

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

Bouncing Dynamics
When we talk about bouncing dynamics, we are referring to how objects like balls move upon impact with surfaces. This motion is more than just a simple bounce; it involves changes in velocity and height after every impact. Various factors affect bouncing dynamics, such as the material of the objects involved, surface texture, and angle of impact.

In our specific exercise, consider that the ball initially dropped from a height will lose some energy in form of sound, heat, and deformation energy upon hitting the plates. This lost energy means the ball doesn’t bounce back to its original height. The analysis of bouncing dynamics involves understanding these losses and how they affect subsequent bounces.

We can conveniently summarize bouncing behavior with a single term: the coefficient of restitution, which we'll explore more in subsequent sections.
Energy Conservation in Collisions
In collisions, especially bouncy ones, energy conservation is key. Although it might seem all kinetic energy would transform back into potential energy, not all of it does.

Upon impact, the ball converts its kinetic energy partly into other forms like thermal energy or sound, hence reducing the potential energy it can reach afterwards.

This is where the coefficient of restitution comes in handy. It's a measure of how well kinetic energy is conserved in a collision. The formula often used \( e = \sqrt{ \frac{h_{f}}{h_{i}} } \) ,
highlights this perfectly: it compares the drop height to the height reached after a bounce. A value of 1 would imply no energy loss (a perfectly elastic collision), while values below 1 indicate energy losses.
Impact Physics
Impact physics delves into the forces, velocities, and energy transfers occurring during the contact between two objects. It's the study of how objects react to the force of impact, which is crucial for understanding any collision event.

Regarding our ball and plates, impact physics helps explain why the coefficient of restitution is vital. It accounts for the nature of the materials and how their inherent properties influence the bounce.
  • The coefficient indicates how 'bouncy' the collision is.
  • Velocity before and after impact changes, and impact physics can compute these variations.


Understanding impact, you start to realize materials (like a hard plate vs. a soft mat) play a huge role in determining the ball's rebounding height and speed. This knowledge is essential when designing materials meant to dampen or enhance bounce.

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Most popular questions from this chapter

A tractor-trailer rig with a 2000-kg tractor, a 4500-kg trailer, and a 3600-kg trailer is traveling on a level road at 90 km/h. The brakes on the rear trailer fail, and the antiskid system of the tractor and front trailer provide the largest possible force that will not cause the wheels to slide. Knowing that the coefficient of static friction is 0.75, determine (a) the shortest time for the rig to a come to a stop, (b) the force in the coupling between the two trailers during that time. Assume that the force exerted by the coupling on each of the two trailers is horizontal.

A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supported by springs. Knowing that the height of the rebound is 0.6 m, determine (a) the velocity of the plate immediately after the impact, (b) the energy lost due to the impact.

A light train made of two cars travels at 45 mi/h. Car A weighs 18 tons, and car B weighs 13 tons. When the brakes are applied, a constant braking force of 4300 lb is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down.

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A 4-lb collar can slide without friction along a horizontal rod and is released from rest at A. The undeformed lengths of springs BA and CA are 10 in. and 9 in., respectively, and the constant of each spring is 2800 lb/in. Determine the velocity of the collar when it has moved 1 in. to the right.

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