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Chapter 8: Problem 3

(a) Show that the kinetic energy \(K\) and the momentum magnitude \(p\) of a particle with mass \(m\) are related by \(K=p^{2} / 2 m .\) (b) A \(0.040-\mathrm{kg}\) cardinal (Richmondena cardinalis) and a \(0.145-\mathrm{kg}\) baseball have the same kinetic energy. Which has the greater magnitude of momentum? What is the ratio of the cardinal's magnitude of momentum to the bascball's? A \(700-N\) man and a \(450-N\) woman have the same momentum. Who has the greater kinetic energy? What is the ratio of the man's kinetic energy to that of the woman?

Short Answer

Expert verified
Baseball has greater momentum; momentum ratio is 0.523. Woman has greater kinetic energy; energy ratio is 0.643.

Step by step solution

01

Understanding the Given Relation

We are provided that kinetic energy \( K \) and momentum \( p \) for a particle of mass \( m \) have the relation \( K = \frac{p^2}{2m} \). This shows that this formula is derived from the definitions of kinetic energy and momentum.
02

Derive Kinetic Energy Formula

The kinetic energy \( K \) of a particle is given by the equation \( K = \frac{1}{2}mv^2 \). We also know that momentum \( p \) is given by \( p = mv \). By substituting \( v = \frac{p}{m} \) into the kinetic energy formula, we get:\[ K = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{p^2}{2m} \]Thus, we've shown the relation \( K = \frac{p^2}{2m} \).
03

Comparing Momentum of Cardinal and Baseball

Both the cardinal (0.040 kg) and the baseball (0.145 kg) have the same kinetic energy. Since \( K = \frac{p^2}{2m} \), for the same \( K \), the momentum \( p = \sqrt{2mK} \). The mass is greater for the baseball, so its \( p \) will be greater as \( p \) is directly proportional to the square root of the mass.
04

Calculating the Ratio of Momenta

The ratio of momentum of the cardinal to the baseball is given by\[\text{Ratio} = \frac{p_{\text{cardinal}}}{p_{\text{baseball}}} = \sqrt{\frac{m_{\text{cardinal}}}{m_{\text{baseball}}}}\]Substituting the masses, we get:\[\text{Ratio} = \sqrt{\frac{0.040}{0.145}} \approx 0.523\]
05

Comparing Kinetic Energy of Man and Woman

A 700-N man and a 450-N woman have equal momentum. Their weight corresponds to mass multiplied by gravity, so \( m_{\text{man}} = \frac{700}{9.8} \) and \( m_{\text{woman}} = \frac{450}{9.8} \). The kinetic energy \( K \) is given by \( K = \frac{p^2}{2m} \), making it inversely proportional to mass for the same momentum. The woman has a smaller mass, thus a higher kinetic energy.
06

Calculating Kinetic Energy Ratio

To find the ratio of the man's kinetic energy to the woman's, use the formula:\[\text{Ratio} = \frac{K_{\text{man}}}{K_{\text{woman}}} = \frac{m_{\text{woman}}}{m_{\text{man}}}\]Substitute the masses:\[\text{Ratio} = \frac{\frac{450}{9.8}}{\frac{700}{9.8}} = \frac{450}{700} \approx 0.643\]

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is a product of an object's mass and its velocity, mathematically expressed as \( p = mv \). This means that heavier objects or those moving faster have greater momentum.
Momentum is particularly vital because it is conserved in closed systems, meaning that the total momentum in a system remains constant if no external forces act on it.
In practical terms, this conservation can help us predict outcomes in collisions and interactions between particles.
  • For objects with the same kinetic energy, those with larger mass will generally have greater momentum.
  • In comparing two objects with the same momentum, the object with greater mass will have lower velocity.
Energy-momentum relation
The energy-momentum relation is crucial for connecting kinetic energy with momentum. It shows how these concepts are intertwined rather than independent.
For non-relativistic physics, the relation is given by \( K = \frac{p^2}{2m} \). This formula comes from substituting the expression for momentum \( p = mv \) into the kinetic energy equation \( K = \frac{1}{2}mv^2 \).
This equation helps to solve problems where either kinetic energy or momentum needs to be determined, given the other.
  • It suggests that for a constant kinetic energy, momentum increases with mass.
  • The expression provides insights into how energy distribution changes with particle mass in particle physics.
Particle physics
Particle physics studies the fundamental particles of the universe and their interactions. In this field, energy and momentum relations become even more significant as particles often move at high speeds near light speed, and traditional assumptions must be adjusted.
In these high-energy scenarios, the full relativistic energy-momentum relationship is used:\[E^2 = (mc^2)^2 + (pc)^2\]This equation links the total energy \( E \) of a particle to its rest mass \( m \) and momentum \( p \).
  • It shows how energy and momentum influence each other and how mass plays a role even when no rest mass exists (as in photons).
  • This relation is fundamental in understanding particle collisions at particle accelerators, where speeds approach light speed.
Mass and Speed Relationship
The relationship between mass and speed is crucial in understanding how objects behave in motion. As an object's speed increases, its kinetic energy also increases exponentially, especially as it approaches relativistic speeds (near the speed of light).
According to classical mechanics, an object with larger mass generally requires more force to accelerate to the same speed as a lighter object. As such, its momentum also rises due to the increased product of mass and speed.
When considering real-world applications:
  • In sports, heavier balls (like the baseball in the exercise) have larger momentum for the same kinetic energy compared to lighter objects (like the cardinal).
  • In transportation or particle physics, adjusting for mass and speed is critical for calculating the energy costs or mechanical requirements involved.

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Most popular questions from this chapter

A 1200 -kg station wagon is moving along a straight highway at 12.0 \(\mathrm{m} / \mathrm{s}\) . Another car, with mass 1800 \(\mathrm{kg}\) and speed 20.0 \(\mathrm{m} / \mathrm{s}\) , has its center of mass 40.0 \(\mathrm{m}\) ahead of the center of mass of the station wagon (Fig. 8.39\() .\) (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (o) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).

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 \(\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\mathrm{i}}\) , (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 \(t=3.0 \mathrm{s} .\)

In a certain men's track and field event, the shotput has a mass of 7.30 \(\mathrm{kg}\) and is released with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) at \(40.0^{\circ}\) above the horizontal over a man's straight left leg. What are the initial horizontal and vertical components of the momentum of this shotput?

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g .\) $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=a t,\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v,\) find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{s}\) . (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m}\) , find the mass of the raindrop at \(t=3.00 \mathrm{s}\) . For many more intriguing aspects of this problem, see \(\mathbf{K} .\) S. Krane, Amer Jour. Phys, Vol. 49\((1981)\) pp. \(113-117\)

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a \(1750-\mathrm{kg}\) car traveling to the right at 1.50 \(\mathrm{m} / \mathrm{s}\) collides with a \(1450-\mathrm{kg}\) car going to the left at 1.10 \(\mathrm{m} / \mathrm{s}\) . Measurements show that the heavier car's speed just after the collision was 0.250 \(\mathrm{m} / \mathrm{s}\) in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.
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