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A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(6.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(3.5 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?

Short Answer

Expert verified
The magnitude of the change in linear momentum is \(6.65 \, \text{kg} \cdot \text{m/s}\).

Step by step solution

01

Understanding Linear Momentum

Linear momentum is given by the formula \( p = mv \), where \( m \) is mass and \( v \) is velocity. The problem involves a ball changing its velocity due to rebounding off a wall, and we are tasked with finding the change in its linear momentum.
02

Calculate Initial Momentum

The initial velocity of the ball is \( 6.0 \, \text{m/s} \), moving towards the wall. Hence, the initial momentum is \( p_i = (0.70 \, \text{kg}) \times (6.0 \, \text{m/s}) = 4.2 \, \text{kg} \cdot \text{m/s} \).
03

Calculate Final Momentum

After rebounding, the ball moves in the opposite direction with a speed of \( 3.5 \, \text{m/s} \). Thus, its final velocity is \( -3.5 \, \text{m/s} \). The final momentum is \( p_f = (0.70 \, \text{kg}) \times (-3.5 \, \text{m/s}) = -2.45 \, \text{kg} \cdot \text{m/s} \).
04

Calculate Change in Momentum

The change in momentum is the final momentum minus the initial momentum: \( \Delta p = p_f - p_i = (-2.45 \, \text{kg} \cdot \text{m/s}) - (4.2 \, \text{kg} \cdot \text{m/s}) \). This simplifies to \( \Delta p = -6.65 \, \text{kg} \cdot \text{m/s} \).
05

Compute Magnitude of Change in Momentum

The magnitude of the change in momentum is the absolute value of \( \Delta p \). Thus, \( |\Delta p| = 6.65 \, \text{kg} \cdot \text{m/s} \).

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Key Concepts

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

Rebounding Ball
When a ball hits a vertical wall and rebounds, a fascinating event occurs in the realm of physics. This process involves rapid deceleration upon impact and then acceleration in the opposite direction. Let's break this down: Imagine you throw a ball towards a wall. As the ball hits the wall, the collision causes a directional change. The ball is momentarily at rest before moving away from the wall. This directional change is the core of what we refer to as a 'rebounding' motion.
Rebounding is significant because it involves changes in energy and momentum. Not all of the initial speed is retained, so post-collision, the ball usually has slower speeds. The amount of energy lost in this process can be due to factors like heat or energy absorbed by the wall.
Understanding rebounding helps us grasp the underlying forces and transformations that occur during such quick and dynamic interactions.
Change in Momentum
Momentum is a fundamental physics concept defined by the product of an object's mass and its velocity. When a rebounding ball is involved, it undergoes a change in momentum. This change in momentum is calculated by looking at the initial and final momenta.
The initial momentum is determined before the ball impacts the wall, and the final momentum after it rebounds. The formula for momentum is clear-cut: \( p = mv \). For our example with the ball, you have:
  • Initial momentum: \( 4.2 \, \text{kg} \cdot \text{m/s} \)
  • Final momentum: \( -2.45 \, \text{kg} \cdot \text{m/s} \)
The change in momentum is the difference between final and initial momentum: \( \Delta p = p_f - p_i \). In our case, this value is \( -6.65 \, \text{kg} \cdot \text{m/s} \). Remember: since momentum has direction, this change accounts for the complete reversal in direction post-collision.
Velocity Calculation
Velocity is crucial in understanding both initial and final movements of the ball. It's the speed of an object in a given direction. When describing the motion of a rebounding ball, both magnitude and direction matter.
Initially, the ball moves towards the wall at \( 6.0 \, \text{m/s} \). After hitting the wall, it reverses direction with a speed of \( 3.5 \, \text{m/s} \) in the opposite direction.
Calculating velocity like this requires attention to direction, not just speed. That's why the final velocity is written as \(-3.5 \, \text{m/s} \), indicating movement in the opposite direction. These calculations help us visualize the motion changes and quantify them accurately.
Conservation of Momentum
The principle of conservation of momentum is pivotal in physics. It states that within a closed system, the total momentum before and after an event remains constant, provided no external forces intervene. Even though a wall interaction might seem like an "external force," for short timescales, we often consider just the system of the object in question.
During a collision, like our rebounding ball, individual momenta change, but total system momentum remains constant. This is because the force exerted on the ball by the wall is equal and opposite to the force exerted on the wall by the ball. The wall absorbs some energy, leading to lower post-collision speeds.
Understanding momentum conservation lets us predict post-collision velocities and analyze complex systems with multiple interacting bodies. This principle gives insight into the motion's predictability and continuity, highlighting the dynamic interplays of forces in motion events.

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Most popular questions from this chapter

In the Olympiad of 708 B.C., some athletes competing in the standing long jump used handheld weights called halteres to lengthen their jumps (Fig. 9-48). The weights were swung up in front just before liftoff and then swung down and thrown backward during the flight. Suppose a modern \(78 \mathrm{~kg}\) long jumper similarly uses two \(5.50 \mathrm{~kg}\) halteres, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. Let his liftoff velocity be \(\vec{v}=(9.5 \hat{i}+4.0 \mathrm{j}) \mathrm{m} / \mathrm{s}\) with or without the halteres, and assume that he lands at the liftoff level. What distance would the use of the halteres add to his range?

After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 \(\mathrm{m}\). During the collision at the bottom of the elevator shaft, a \(90 \mathrm{~kg}\) passenger is stopped in \(5.0 \mathrm{~ms}\). (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

In the ammonia \(\left(\mathrm{NH}_{3}\right)\) molecule of Fig. 9-29, three hydrogen (H) atoms form an equilateral triangle, with the center of the triangle at distance \(d=\) \(9.40 \times 10^{-11} \mathrm{~m}\) from each hydrogen atom. The nitrogen \((\mathrm{N})\) atom is at the apex of a pyramid, with the three hydrogen atoms forming the base. The nitrogen-to-hydrogen atomic mass ratio is \(13.9\), and the nitrogen-tohydrogen distance is \(L=10.14 \times\) \(10^{-11} \mathrm{~m}\). What are the (a) \(x\) and (b) \(y\) coordinates of the molecule's center of mass?

A ball of mass \(50 \mathrm{~g}\) moving with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) strikes a wall at an angle of incidence \(45^{\circ}\) and is reflected from the wall at the same angle and with the same speed. See the overhead view in Fig. 9-38. Calculate (a) the magnitude of the change \(\Delta \vec{p}\) in the momentum of the ball, (b) the change in the magnitude of the momentum \(\vec{p}\) of the ball, and (c) the change in the magnitude of the momentum of the wall.

A particle of mass \(2.0 \mathrm{~m}\) is projected at an angle of \(45^{\circ}\) with the horizontal with a speed of \(20 \sqrt{2} \mathrm{~m} / \mathrm{s}\). After \(1.0 \mathrm{~s}\), an explosion occurs and the particle is broken into two equal pieces. One piece is momentarily at rest before it falls. Find the maximum height attained by the other piece.

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