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Chapter 7: Problem 4

A \(0.70 \mathrm{~kg}\) ball is moving horizontally with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) when it strikes a vertical wall. The ball rebounds with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in translational momentum of the ball?

Short Answer

Expert verified
4.9 \mathrm{~kg \, m/s}

Step by step solution

01

- Identify Given Values

Identify the given values from the problem. Here, the mass of the ball is given as \(0.70 \mathrm{~kg}\). The initial speed (before it strikes the wall) is \(5.0 \mathrm{~m/s}\) and the final speed (after rebounding) is \(2.0 \mathrm{~m/s}\).
02

- Calculate Initial Momentum

The formula for momentum is given by \( p = mv \), where \( m \) is mass and \( v \) is velocity. Calculate the initial momentum of the ball: \[ p_{initial} = 0.70 \mathrm{~kg} \times 5.0 \mathrm{~m/s} = 3.5 \mathrm{~kg \, m/s} \]
03

- Calculate Final Momentum

Similarly, calculate the final momentum, but keep in mind that the ball is rebounding in the opposite direction. If we take the initial direction as positive, the final velocity will be negative. Thus, \[ p_{final} = 0.70 \mathrm{~kg} \times (-2.0 \mathrm{~m/s}) = -1.4 \mathrm{~kg \, m/s} \]
04

- Compute the Change in Momentum

The change in momentum (\( \triangle p \)) is given by the difference between the final and initial momentum: \[ \triangle p = p_{final} - p_{initial} = -1.4 \mathrm{~kg \, m/s} - 3.5 \mathrm{~kg \, m/s} \] Simplify the equation: \[ \triangle p = -4.9 \mathrm{~kg \, m/s} \]
05

- Find the Magnitude of the Change in Momentum

The magnitude of \( \triangle p \) is the absolute value of the change in momentum: \[ | \triangle p | = 4.9 \mathrm{~kg \, m/s} \]

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Key Concepts

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

Translational Momentum
Translational momentum is a measure of the motion of an object moving in a straight line. It is calculated using the formula \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In this problem, the ball demonstrates translational momentum as it moves horizontally towards and then away from the wall. Understanding this formula helps us determine the initial and final momentum of the ball before and after it hits the wall. This concept is fundamental in analyzing the ball's motion and finding the change in momentum.
Conservation of Momentum
The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In the scenario given, even though the ball's direction changes after hitting the wall, the overall momentum before and after the impact remains balanced in our closed system. This implies that any change in the ball's momentum directly corresponds to the forces involved during the collision. To apply this principle, we need to calculate the initial and final momentum and compare them to find the change or difference.
Momentum Calculation
To calculate momentum, we use the formula \( p = mv \). Let's break down the calculations from the exercise:
\( p_{initial} = 0.70 \text{ kg} \times 5.0 \text{ m/s} = 3.5 \text{ kg m/s} \).
Here, we multiply the mass of the ball by its initial velocity.
Next, we calculate the final momentum by considering the direction change:
\( p_{final} = 0.70 \text{ kg} \times (-2.0 \text{ m/s}) = -1.4 \text{ kg m/s} \).
It's essential to account for direction, especially since momentum is a vector quantity.
Initial and Final Velocities
Understanding initial and final velocities is crucial for momentum problems. Here, the ball has an initial velocity of 5.0 m/s before striking the wall and a final velocity of 2.0 m/s after rebounding.
Note the rebound indicates a change in direction, making the final velocity negative compared to the initial direction. Properly marking these velocities as positive or negative ensures accuracy in calculating momentum changes. In our problem, the initial and final velocities help us determine the magnitude and direction of momentum.
Absolute Value in Physics
The absolute value is used to find the magnitude of quantities regardless of their direction. In physics, we often focus on the magnitude, especially when dealing with changes in vector quantities like momentum.
In our exercise, the change in momentum is found by subtracting the initial momentum from the final momentum: \([-4.9 \text{ kg m/s} = -1.4 \text{ kg m/s} - 3.5 \text{ kg m/s}] \).
The magnitude of the change in momentum is the absolute value: \([| \triangle p | = 4.9 \text{ kg m/s}] \).
This step ensures that regardless of the direction, we have a clear measure of the change in momentum.

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Most popular questions from this chapter

The principle of conservation of momentum is useful in some situations and not in others. Describe how you obtain the impulse-momentum theorem from Newton's Second Law and what situations lead to momentum conservation. How would you decide whether conservation of momentum could be useful in a particular problem?

A professor of physics is going ice skating for the first time. He has gotten himself into the middle of an ice rink and cannot figure out how to make the skates work. Every motion he makes simply causes his feet to slip on the ice and leaves him in the same place he started. He decides that he can get off the ice by throwing his gloves in the opposite direction. (a) Suppose he has a mass \(M\) and his gloves have a mass \(m\). If he throws the gloves as hard as he can away from him, they leave his hand with a velocity \(\vec{v}_{\text {glove }}\) Explain whether or not he will move. If he does move, calculate his velocity, \(\vec{v}_{\text {prof }}\) (b) Discuss his motion from the point of view of the forces acting on him. (c) If the ice rink is \(10 \mathrm{~m}\) in diameter and the skater starts in the center, estimate how long it will take him to reach the edge, assuming there is no friction at all.

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2}\) ?

A \(0.30 \mathrm{~kg}\) softball has a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a) \(20 \mathrm{~m} / \mathrm{s}\), vertically downward and (b) \(20 \mathrm{~m} / \mathrm{s}\), horizontally away from the batter and back toward the pitcher?
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