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(II) The momentum of a particle, in SI units, is given by \(\overrightarrow{\mathbf{p}}=\) \(4.8 t^{2} \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}}-8.9 t \hat{\mathbf{k}} .\) What is the force as a function of time?

Short Answer

Expert verified
The force as a function of time is \( \overrightarrow{\mathbf{F}} = 9.6 t \hat{\mathbf{i}} - 8.9 \hat{\mathbf{k}} \).

Step by step solution

01

Understand the Relationship Between Force and Momentum

The force acting on a particle can be derived from the time derivative of its momentum. We use Newton's second law, which states that the force is equal to the rate of change of momentum: \[ \overrightarrow{\mathbf{F}} = \frac{d\overrightarrow{\mathbf{p}}}{dt} \] Identifying this relationship allows us to find the force as a function of time.
02

Differentiate Each Component of the Momentum Vector

The given momentum vector is \( \overrightarrow{\mathbf{p}}= 4.8 t^{2} \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} - 8.9 t \hat{\mathbf{k}} \). To find the force, differentiate each component with respect to time:- For the \( \hat{\mathbf{i}} \) component: \( \frac{d}{dt}(4.8 t^2) = 9.6 t \).- For the \( \hat{\mathbf{j}} \) component: \( \frac{d}{dt}(-8.0) = 0 \) since it's a constant.- For the \( \hat{\mathbf{k}} \) component: \( \frac{d}{dt}(-8.9 t) = -8.9 \).
03

Combine the Components into the Force Vector

Combine the differentiated components to form the complete force vector:\[ \overrightarrow{\mathbf{F}} = 9.6 t \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 8.9 \hat{\mathbf{k}} \]Thus, the force as a function of time is a vector given by these components.

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Key Concepts

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

Newton's Second Law
Newton’s second law is a core principle in physics describing how the velocity of an object changes when it is subject to an external force. This law is elegantly formulated as \( \overrightarrow{\mathbf{F}} = \frac{d\overrightarrow{\mathbf{p}}}{dt} \), which means that the force \( \overrightarrow{\mathbf{F}} \) is equal to the rate of change of momentum \( \overrightarrow{\mathbf{p}} \). In simpler terms, momentum, represented by \( \overrightarrow{\mathbf{p}} \), is the product of an object's mass and its velocity. Newton's second law tells us that the force exerted on an object causes its momentum to change over time.

For instance, if a particle's momentum is changing due to a varying external force, differentiating the momentum with respect to time gives us the force acting on the particle. This concept is essential for understanding how different forces can affect the motion of particles in various directions.
Differentiation
Differentiation is a mathematical process used to determine the rate at which a quantity changes with respect to another. In our exercise, differentiation is used to find the force from the given momentum vector. By applying this operation, you can calculate the instantaneous rate of change, which in the context of vectors, informs us how each component of the momentum is changing over time.

**Steps to Differentiate:**
  • Identify each term of the vector that needs differentiation.
  • Apply the derivative to each term individually based on its mathematical expression.
  • Understand that the differentiation of a constant yields zero, revealing no change over time.

For example, if a term in the momentum vector is \(4.8 t^2\), differentiating with respect to time \(t\) results in \(9.6 t\), illustrating how rapidly that component changes. This process is key to transforming momentum expressions into force vectors.
Vector Components
Vector components are the building blocks of any vector, representing its influence in different directions in space. In a three-dimensional space, we typically break down a vector into its \( \hat{\mathbf{i}} \), \( \hat{\mathbf{j}} \), and \( \hat{\mathbf{k}} \) components along the x, y, and z axes respectively. These components make it easier to perform mathematical operations, such as differentiation.

**Understanding Components:**
  • Each component corresponds to a specific direction's influence.
  • Components can be differentiated individually, simplifying complex calculations.
  • Combining the effects of each component gives a clear picture of the total vector force or momentum.

For the given momentum vector \( \overrightarrow{\mathbf{p}} = 4.8 t^{2} \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} - 8.9 t \hat{\mathbf{k}} \), each component represents how the particle moves or changes in a particular direction. Once differentiated, these components align to form the force vector, \( \overrightarrow{\mathbf{F}} = 9.6t \hat{\mathbf{i}} + 0 \hat{\mathbf{j}} - 8.9 \hat{\mathbf{k}} \), indicating the acting force in each direction.

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Most popular questions from this chapter

(II) A \(925-\mathrm{kg}\) two-stage rocket is traveling at a speed of \(6.60 \times 10^{3} \mathrm{~m} / \mathrm{s}\) away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of \(2.80 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to each other along the original line of motion. \((a)\) What is the speed and direction of each section (relative to Earth) after the explosion? (b) How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

Two astronauts, one of mass \(65 \mathrm{~kg}\) and the other \(85 \mathrm{~kg}\), are initially at rest in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved \(12 \mathrm{~m} ?\)

(II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy \((\Delta K / K)\) gets transferred to the second ball?

(II) A 0.145-kg baseball pitched horizontally at \(32.0 \mathrm{~m} / \mathrm{s}\) strikes a bat and is popped straight up to a height of \(36.5 \mathrm{~m}\). If the contact time between bat and ball is \(2.5 \mathrm{~ms},\) calculate the average force between the ball and bat during contact.

(II) A mass \(m_{\mathrm{A}}=2.0 \mathrm{~kg},\) moving with velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}=\) \((4.0 \hat{\mathbf{i}}+5.0 \hat{\mathbf{j}}-2.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s},\) collides with mass \(m_{\mathrm{B}}=3.0 \mathrm{~kg}\) which is initially at rest. Immediately after the collision, mass \(m_{\mathrm{A}}\) is observed traveling at velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}^{\prime}=(-2.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s}\) Find the velocity of mass \(m_{\mathrm{B}}\) after the collision. Assume no outside force acts on the two masses during the collision.

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