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(I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s?

Short Answer

Expert verified
The magnitude of the momentum of the sparrow is 0.2352 kg·m/s.

Step by step solution

01

Understand the Formula for Momentum

Momentum, represented by the symbol \( p \), is calculated using the formula \( p = mv \), where \( m \) is the mass of the object in kilograms and \( v \) is its velocity in meters per second.
02

Convert Mass to Kilograms

The mass of the sparrow is given as 28 grams. Convert this mass to kilograms by dividing by 1000, as there are 1000 grams in a kilogram: \( m = \frac{28}{1000} = 0.028 \text{ kg} \).
03

Use the Momentum Formula

Substitute the values of mass \( m = 0.028 \text{ kg} \) and speed \( v = 8.4 \text{ m/s} \) into the formula \( p = mv \): \( p = 0.028 \times 8.4 \).
04

Calculate the Momentum

Calculate \( 0.028 \times 8.4 \) to find the momentum: \( p = 0.2352 \text{ kg} \cdot \text{m/s} \).
05

Conclude the Solution

The magnitude of the momentum of the sparrow is \( 0.2352 \text{ kg} \cdot \text{m/s} \).

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Key Concepts

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

Mass Conversion
In physics, mass is often represented in different units depending on the context. To work with the momentum equation, we need the mass in kilograms. This is crucial because the standard unit of mass in the International System of Units (SI) is the kilogram. When given a mass in grams, as in the problem of our sparrow, it's important to convert it to kilograms.
  • To convert grams to kilograms, divide by 1000, since 1000 grams equal 1 kilogram.
For example, the sparrow's mass is 28 grams. To convert this to kilograms, you perform the calculation: \[\text{Mass in kilograms} = \frac{28 \text{ grams}}{1000} = 0.028 \text{ kg}\] This conversion ensures that the units are consistent and allows you to accurately apply the momentum formula.
Momentum Formula
Momentum is a measure of how much motion an object has and is represented as a product of mass and velocity. The formula for momentum is expressed as: \[p = mv\]- **\(p\)** stands for momentum- **\(m\)** is mass in kilograms- **\(v\)** is velocity in meters per secondThis formula shows that momentum depends directly on both the mass of the object and its velocity. Larger mass or higher velocity means greater momentum. It's important to ensure that both mass and velocity are in their correct units (kilograms and meters per second respectively), making calculations straightforward.
For our sparrow, substituting the known values into the equation gives us:\[p = 0.028 \text{ kg} \times 8.4 \text{ m/s}\]This helps us find the momentum by performing a simple multiplication of the two values.
Velocity Calculation
Velocity refers to the speed of an object in a specific direction and is a crucial component in calculating momentum. The sparrow's velocity in this exercise is given directly as 8.4 meters per second. Here are key points to consider:
  • Velocity is directional, meaning it specifies not only how fast something is moving, but in which direction.
  • However, in this problem, we focus on speed alone as we're calculating the magnitude of momentum, which is why direction isn't considered.
In cases where velocity isn't given, it might be necessary to calculate it using additional data. However, in situations like this exercise, simply plugging in the given velocity into the momentum equation becomes straightforward.
By understanding velocity's role in momentum, we can appreciate why it's vital to have accurate speed values in calculations involving dynamic movements.

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Most popular questions from this chapter

(II) An atomic nucleus at rest decays radioactively into an alpha particle and a different nucleus.What will be the speed of this recoiling nucleus if the speed of the alpha particle is \(2.8 \times 10^5 m/s\)? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle.

(II) Billiard ball A of mass \(m_A = 0.120 kg\) moving with speed \(v_A = 2.80 m/s\) strikes ball B, initially at rest, of mass \(m_B = 0.140 kg\). As a result of the collision, ball A is deflected off at an angle of 30.0\(^{\circ}\) with a speed \(\upsilon'_{A} = 2.10 m/s\). \((a)\) Taking the \(\chi\) axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the \(\chi\) and \(y\) directions separately. (b) Solve these equations for the speed \(\upsilon'_{B}\), and angle \(\theta'_{B}\), of ball B after the collision. Do not assume the collision is elastic.

(II) A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum's maximum displacement.

(II) The masses of the Earth and Moon are \(5.98 \times 10^{24} kg\) and \(7.35 \times 10^{22} kg\), respectively, and their centers are separated by \(3.84 \times 10^8 m\). \((a)\) Where is the \(_{CM}\) of the Earth-Moon system located? \((b)\) What can you say about the motion of the Earth-Moon system about the Sun, and of the Earth and Moon separately about the Sun?

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