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Chapter 9: Problem 17

A 15.0-g marble is dropped from rest onto the floor \(1.44 \mathrm{m}\) below. (a) If the marble bounces straight upward to a height of \(0.640 \mathrm{m},\) what are the magnitude and direction of the impulse delivered to the marble by the floor? (b) If the marble had bounced to a greater height, would the impulse delivered to it have been greater or less than the impulse found in part (a)? Explain.

Short Answer

Expert verified
Impulse is 0.1329 Ns upward; a higher bounce means a greater impulse.

Step by step solution

01

Understanding the problem

The marble is dropped from a height of 1.44 meters, bounces back to a height of 0.640 meters. We need to find the impulse imparted by the floor when it hits and bounces back, and how a higher bounce would affect this impulse.
02

Calculate initial velocity before impact

The marble is initially dropped from rest, hence its initial velocity is 0 m/s. On impact with the floor, the velocity before impact can be calculated using the equation of motion: \[ v_i = \sqrt{2gh} \]where \(g = 9.8 \text{ m/s}^2\) is the acceleration due to gravity and \(h = 1.44 \text{ m} \). This gives:\[ v_i = \sqrt{2 \times 9.8 \times 1.44} \approx 5.33 \text{ m/s} \]
03

Calculate velocity after bounce

Post-bounce, the velocity can be calculated for the marble to reach a height of 0.640 meters using the same equation:\[ v_f = \sqrt{2gh'} \]with \( h' = 0.640 \text{ m} \), we have:\[ v_f = \sqrt{2 \times 9.8 \times 0.640} \approx 3.54 \text{ m/s} \]
04

Determining the direction of velocities

The initial velocity prior to impact with the floor is downward (hence negative) and calculated as \( v_i = -5.33 \text{ m/s} \). The post-bounce velocity is upward and is positive with \( v_f = 3.54 \text{ m/s} \).
05

Calculate the impulse delivered

Impulse \( J \) is given by the change in momentum:\[ J = m(v_f - v_i) \]Substitute the values: mass \( m = 15 \text{ g} = 0.015 \text{ kg} \), \( v_f = 3.54 \text{ m/s} \), and \( v_i = -5.33 \text{ m/s} \):\[ J = 0.015 \times (3.54 - (-5.33)) = 0.1329 \text{ Ns} \]This is directed upward.
06

Analysis of a higher bounce

If the marble bounces higher, the final velocity required to reach that height is greater. This leads to a larger change in momentum and thus a greater impulse. Therefore, a higher bounce implies that a greater impulse is delivered.

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Key Concepts

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

Projectile Motion
When we talk about projectile motion, we are discussing the movement of an object that is thrown, dropped, or propelled along a trajectory under the influence of gravity. In the scenario involving a marble being dropped, the initial motion is vertical and influenced solely by gravity.

Important aspects of projectile motion include:
  • Initial velocity: At the start, our marble was released from a height of 1.44 meters with an initial velocity of 0 m/s.
  • Acceleration due to gravity: Gravity accelerates the marble downward at 9.8 m/s虏.
  • Final velocity before impact: Calculated using the motion equation, resulting in a velocity of approximately 5.33 m/s just before it hits the floor.
Understanding these factors helps us calculate how the marble moves and at what speed it strikes the ground. Once we find the velocity upon impact, we can predict its behavior post-bounce鈥攍ike reaching a height of 0.640 meters, which signifies its projectile path in the opposite direction.
Conservation of Energy
The conservation of energy principle tells us that energy cannot be created or destroyed, only transformed from one form to another. Let's see how this applies to our marble exercise:

As the marble falls:
  • Potential Energy (PE) Conversion: Initially, when the marble is at 1.44 meters, it possesses gravitational potential energy, calculated as \( PE = mgh \), where \( m \) is mass, \( g \) is gravity, and \( h \) is height.
  • Kinetic Energy (KE) Conversion: As the marble falls, its potential energy transforms into kinetic energy (\( KE = \frac{1}{2}mv^2 \)), contributing to the velocity gained upon hitting the ground.
When the marble bounces back, some kinetic energy transforms back into potential energy, enabling it to reach 0.640 meters of height. The energy transformation during the bounce wasn't totally efficient due to energy losses possibly from air resistance and heat upon impact, illustrated by the lower rebound height.
Newton's Laws of Motion
Newton's Laws of Motion are fundamental to understanding the movement of the marble after collision with the floor.

First law (Inertia): The marble continues in its state of motion or rest unless acted upon by another force. Dropped from rest, the marble continues to accelerate until hitting the floor, where the force from the floor (impulse) changes its state into motion upwards.
  • Second Law (Force and Acceleration): Expressed as \( F = ma \), this law relates to how forces interact with inertia and acceleration. During impact, the force from the floor applies an impulse changing the marble鈥檚 momentum.
  • Third Law (Action and Reaction): It states that for every action, there's an equal and opposite reaction. When the marble exerts a force on the floor, the floor exerts an equal force back (impulse), resulting in movement upward with velocity \( v_f = 3.54 \text{ m/s} \).
These laws allow us to calculate the impulse delivered by using momentum change derived from the initial and final velocities of the marble. They offer a vivid explanation of how the floor's reaction propels the marble back upwards, corresponding to the height it eventually reaches.

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