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CALC Starting at \(t=0,\) a horizontal net force \(\vec{F}=\) \((0.280 \mathrm{N} / \mathrm{s}) \hat{t} \hat{\imath}+\left(-0.450 \mathrm{N} / \mathrm{s}^{2}\right) t^{2} \hat{\mathrm{J}}\) is applied to a box that has an initial momentum \(\vec{p}=(-3.00 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\imath}+(4.00 \mathrm{kg} \cdot \) \(\mathrm{m} / \mathrm{s} ) \hat{\boldsymbol{J}}\) . What is the momentum of the box at \(t=2.00 \mathrm{s} ?\)

Short Answer

Expert verified
The momentum of the box at \(t=2\,\text{s}\) is \((-2.44\,\text{kg}\cdot\text{m/s}) \hat{\imath} + (2.80\,\text{kg}\cdot\text{m/s}) \hat{\jmath}\).

Step by step solution

01

Understand the Net Force

The net force applied to the box is given as \( \vec{F}(t) = (0.280\, \text{N/s})t \hat{\imath} + (-0.450\, \text{N/s}^2)t^2 \hat{\jmath} \). It varies with time, so we need to integrate it over time to find the change in momentum.
02

Use Impulse-Momentum Theorem

According to the impulse-momentum theorem, the change in momentum \( \Delta \vec{p} \) is equal to the integral of force with respect to time, i.e., \( \Delta \vec{p} = \int_{0}^{2} \vec{F}(t) \, dt \).
03

Set Up the Integral for \( \hat{\imath} \) Component

For the \( \hat{\imath} \) component: \( \Delta p_{i} = \int_{0}^{2} (0.280t) \, dt \). This simplifies to \( 0.280 \int_{0}^{2} t \, dt \).
04

Calculate the \( \hat{\imath} \) Component Change in Momentum

Solve the integral: \[ \Delta p_{i} = 0.280 \left[ \frac{t^2}{2} \right]_{0}^{2} = 0.280 \left( \frac{4}{2} - 0 \right) = 0.280 \times 2 = 0.560 \text{ kg} \cdot \text{m/s}. \]
05

Set Up the Integral for \( \hat{\jmath} \) Component

For the \( \hat{\jmath} \) component: \( \Delta p_{j} = \int_{0}^{2} (-0.450 t^2) \, dt \). This simplifies to \( -0.450 \int_{0}^{2} t^2 \, dt \).
06

Calculate the \( \hat{\jmath} \) Component Change in Momentum

Solve the integral: \[ \Delta p_{j} = -0.450 \left[ \frac{t^3}{3} \right]_{0}^{2} = -0.450 \left( \frac{8}{3} \right) = -0.450 \times \frac{8}{3} = -1.200 \text{ kg} \cdot \text{m/s}. \]
07

Determine Final Momentum

The initial momentum is given by \( \vec{p}_{initial} = (-3.00\, \text{kg}\cdot\text{m/s}) \hat{\imath} + (4.00\, \text{kg}\cdot\text{m/s}) \hat{\jmath} \). Add the change in momentum to the initial momentum: \[ \vec{p}_{final} = \vec{p}_{initial} + \Delta \vec{p} = (-3.00 + 0.560) \hat{\imath} + (4.00 - 1.200) \hat{\jmath}. \]
08

Calculate the Final Momentum

Calculate each component: \( \hat{\imath} \) component: \(-3.00 + 0.560 = -2.44\, \text{kg}\cdot\text{m/s} \) \( \hat{\jmath} \) component: \(4.00 - 1.200 = 2.80\, \text{kg}\cdot\text{m/s} \) Thus, \( \vec{p}_{final} = (-2.44\, \text{kg}\cdot\text{m/s}) \hat{\imath} + (2.80\, \text{kg}\cdot\text{m/s}) \hat{\jmath}. \)

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Key Concepts

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

Impulse-Momentum Theorem
The impulse-momentum theorem tells us an important fact about force and motion. According to this theorem, the change in an object's momentum is equal to the impulse applied to it. Impulse, in turn, is the integral of the net force over time.
This means that if you apply a force to an object for a certain amount of time, you change its momentum based on the size of the force and how long it's applied. Mathematically, \[ \Delta \vec{p} = \int \vec{F}(t) \, dt \]
In our exercise, the net force is time-dependent, so we integrate it over our desired time frame to find out how much the momentum has changed.
  • Impulse represents the area under a force-time graph.
  • Momentum depends on both the mass and velocity of an object, so it's a vector quantity.
This theorem is crucial for connecting force and motion parameters in dynamic systems.
Integral Calculus in Physics
Integral calculus is fundamental in physics to calculate quantities that accumulate over time, like displacement or change in momentum. In the impulse-momentum theorem, we use integral calculus to compute the impulse.
We come across integrals that represent the area under a curve on a graph, which in this context relates to force versus time or acceleration.
For our exercise, we need to integrate the given force function over time from 0 to 2 seconds to find changes in the different components of momentum.
  • We compute the definite integrals for each component of force - the results give the changes in the respective components of momentum.
  • The integral helps us find accumulated quantities over time, making it ideal for dynamic analyses.
  • In vector calculus, each component (such as \(\hat{\imath}\) or \(\hat{\jmath}\)) is treated separately, applying integral calculus to each individually.
This use of integrals allows us to link force-time data directly with momentum changes over time, providing deeper insight into evolving systems.
Vector Components
When dealing with physics problems involving vectors, breaking them into their components is key. Vectors like force and momentum have magnitude and direction, often represented in three axes: \(\hat{\imath}\), \(\hat{\jmath}\), and \(\hat{k}\).
In our exercise, all motion is horizontal and vertical, so we only consider the \(\hat{\imath}\) and \(\hat{\jmath}\) components.
To solve for momentum, each component is calculated separately:
  • Vector components allow for simpler mathematical manipulation and integration.
  • This approach helps tackle forces or motions acting in different directions independently.
By treating the \(\hat{\imath}\) and \(\hat{\jmath}\) directions separately, we apply force laws to each axis, then combine results to form the total momentum vector.
Net Force Analysis
Understanding net force is crucial for analyzing motion. Net force is the vector sum of all forces acting on an object. It dictates how an object moves according to Newton's Second Law of Motion.
In our problem, the net force changes over time, which influences how quickly or slowly the box's momentum changes.
Key points about net force include:
  • It determines the acceleration and subsequently the change in velocity and momentum of a body.
  • Time-varying forces, like in our exercise, require calculus to precisely describe their effect over a specific duration.
  • By integrating these forces, we derive changes in momentum, essential for dynamics predictions.
Performing net force analysis provides insight into real-world applications, predicting how multiple forces combine to affect an object's motion.

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Most popular questions from this chapter

An astronaut in space cannot use a conventional means, such as a scale or balance, to determine the mass of an object. But she does have devices to measure distance and time accurately. She knows her own mass is 78.4 \(\mathrm{kg}\) , but she is unsure of the mass of a large gas canister in the airless rocket. When this canister is approaching her at 3.50 \(\mathrm{m} / \mathrm{s}\) , she pushes against it, which slows it down to 1.20 \(\mathrm{m} / \mathrm{s}\) (but does not reverse it) and gives her a speed of 2.40 \(\mathrm{m} / \mathrm{s} .\) What is the mass of this canister?

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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400 -kg ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?

CALC A system consists of two particles. At \(t=0\) one particle is at the origin; the other, which has a mass of \(0.50 \mathrm{kg},\) is on the \(y\) -axis at \(y=6.0 \mathrm{m} .\) At \(t=0\) the center of mass of the system is on the \(y\) -axis at \(y=2.4 \mathrm{m} .\) The velocity of the center of mass is given by \(\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\imath}\) (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time \(t .\) (c) Find the net external force acting on the system at \(t=3.0 \mathrm{s}\) .

A steel ball with mass 40.0 \(\mathrm{g}\) is dropped from a height of 2.00 \(\mathrm{m}\) onto a horizontal steel slab. The ball rebounds to a height of 1.60 \(\mathrm{m} .\) (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 \(\mathrm{ms}\) , find the average force on the ball during impact.

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