# 力学和质代写|Mechanics and Matter PHYS10006 University of Bristol Assignment

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Mechanics and Matter PHYS10006力学和质代写代考辅导服务！

## Instructions:

Mechanics is a branch of physics that studies the motion of objects and the forces that cause them. It is concerned with how objects move and interact with each other, and how to describe and predict these behaviors using mathematical models. Mechanics is one of the oldest and most fundamental branches of physics, and has many important applications in engineering, technology, and everyday life.

Matter is anything that has mass and takes up space. It includes all the substances that we can see and touch, as well as those that are invisible to us, such as gases and subatomic particles. Matter is made up of tiny particles called atoms, which are themselves composed of even smaller particles such as protons, neutrons, and electrons.

Mechanics and matter are closely related, since mechanics is concerned with the motion and behavior of objects, and all objects are made of matter. The behavior of matter can be described and predicted using the laws of mechanics, which tell us how objects move and interact with each other under various conditions. For example, mechanics can be used to describe the motion of a ball rolling down a hill, the behavior of a spring as it is compressed and released, or the movement of planets and stars in the universe.

In summary, mechanics and matter are two important concepts in physics that are closely related. Mechanics helps us understand how objects move and interact with each other, while matter is the substance that makes up these objects and is affected by these interactions. Together, these concepts help us to understand the behavior of the physical world around us.

In a crash test, a truck with mass $2500 \mathrm{~kg}$ traveling at $24 \mathrm{~m} / \mathrm{s}$ smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is $0.72 \mathrm{~m}$ shorter than before. (b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)

We can use the conservation of energy to find the initial kinetic energy of the truck:

$\frac{1}{2} m v^2=\frac{1}{2}(2500 \mathrm{~kg})(24 \mathrm{~m} / \mathrm{s})^2=1.44 \times 10^6 \mathrm{~J}$

During the collision, this kinetic energy is converted into deformation work done on the truck, which we can calculate from the change in length:

$W=\frac{1}{2} k x^2=\frac{1}{2} \frac{F}{\Delta x} x^2$

where $k$ is the spring constant of the deformed truck, $x$ is the amount of deformation, and $F$ is the average force exerted by the wall on the truck. Since the truck doesn’t rebound, we can assume that all of the initial kinetic energy is converted into deformation work, so we can equate these two expressions and solve for $F$:

$1.44 \times 10^6 \mathrm{~J}=\frac{1}{2} \frac{F}{0.72 \mathrm{~m}}(0.72 \mathrm{~m})^2 \Rightarrow F=2.96 \times 10^6 \mathrm{~N}$

(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?

We can use the impulse-momentum theorem to find the time interval $\Delta t$ during which the force is exerted:

$F \Delta t=\Delta p=m v_f-m v_i=2500 \mathrm{~kg}(0 \mathrm{~m} / \mathrm{s}-24 \mathrm{~m} / \mathrm{s})=-6.0 \times 10^4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Since the truck comes to a stop, the change in momentum is negative. Solving for $\Delta t$ gives:

$\Delta t=\frac{-6.0 \times 10^4 \mathrm{~kg} \mathrm{~m} / \mathrm{s}}{2.96 \times 10^6 \mathrm{~N}}=0.020 \mathrm{~s}$

So the collision lasts about 0.020 seconds.

It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force $m g$ on the truck. This large ratio shows why a collision is so damaging.

The weight of the truck is $mg = (2500 \mathrm{~kg})(9.81 \mathrm{~m/s^2}) = 24.5 \times 10^3 \mathrm{~N}$. The ratio of the force of the wall to the gravitational force on the truck is:

$\frac{2.96 \times 10^6 \mathrm{~N}}{24.5 \times 10^3 \mathrm{~N}}=120$

So the force exerted by the wall is 120 times greater than the weight of the truck. This large ratio shows why a collision can be so damaging.

# 机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment

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Assignment-daixieTM为您提供卡迪夫大学Cardiff University PX1121 Mechanics and Matter代写代考辅导服务！

## Instructions:

Mechanics is the branch of physics that deals with the motion of objects under the influence of forces. It is concerned with describing and predicting the behavior of physical systems, including both macroscopic objects like cars and airplanes, as well as microscopic particles like atoms and molecules.

Matter, on the other hand, refers to anything that has mass and takes up space. It is the basic building block of the universe, and includes everything from tiny subatomic particles to massive planets and stars.

Mechanics and matter are intimately connected, as the behavior of matter is governed by the laws of mechanics. For example, the motion of a ball thrown into the air can be described using the principles of mechanics, as can the behavior of atoms and molecules in a gas or liquid.

Understanding the relationship between mechanics and matter is crucial for many areas of science and technology, including engineering, materials science, and many others.

Surface tension: Thermodynamic properties of the interface between two phases are described by a state function called the surface tension $\mathcal{S}$. It is defined in terms of the work required to increase the surface area by an amount $d A$ through $d W=\mathcal{S} d A$.
(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius $R$ is larger than outside pressure by $2 \mathcal{S} / R$. What is the air pressure inside a soap bubble of radius $R$ ?

The work done by a water droplet on the outside world, needed to increase the radius from $R$ to $R+\Delta R$ is
$$\Delta W=\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R$$
where $P$ is the pressure inside the drop and $P_o$ is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy $\mathcal{S} \Delta A=\mathcal{S} \cdot 8 \pi R \cdot \Delta R$, where $\mathcal{S}$ is the surface tension, and
$$\Delta W_{\text {total }}=0, \Longrightarrow \Delta W_{\text {pressure }}=-\Delta W_{\text {surface }},$$
resulting in
$$\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R=\mathcal{S} \cdot 8 \pi R \cdot \Delta R, \quad \Longrightarrow \quad\left(P-P_o\right)=\frac{2 \mathcal{S}}{R}$$
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and
$$P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R}$$
$$P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R}$$
Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.

(b) A water droplet condenses on a solid surface. There are three surface tensions involved $\mathcal{S}{a w}, \mathcal{S}{s w}$, and $\mathcal{S}_{s a}$, where $a, s$, and $w$ refer to air, solid and water respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting).

When steam condenses on a solid surface, water either forms a droplet, or spreads on the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore
$$d E=S_{a w} d A_{a w}+S_{a s} d A_{a s}+S_{w s} d A_{w s}=0 .$$

Since the total surface area of the solid is constant,
$$d A_{a s}+d A_{w s}=0 .$$
From geometrical considerations (see proof below), we obtain
$$d A_{w s} \cos \theta=d A_{a w} .$$
From these equations, we obtain
$$d E=\left(S_{a w} \cos \theta-S_{a s}+S_{w s}\right) d A_{w s}=0, \quad \Longrightarrow \quad \cos \theta=\frac{S_{a s}-S_{w s}}{S_{a w}} .$$
Proof of $d A_{w s} \cos \theta=d A_{a w}$ : Consider a droplet which is part of a sphere of radius $R$, which is cut by the substrate at an angle $\theta$. The areas of the involved surfaces are
$$A_{w s}=\pi(R \sin \theta)^2, \quad \text { and } \quad A_{a w}=2 \pi R^2(1-\cos \theta) .$$
Let us consider a small change in shape, accompanied by changes in $R$ and $\theta$. These variations should preserve the volume of water, i.e. constrained by
$$V=\frac{\pi R^3}{3}\left(\cos ^3 \theta-3 \cos \theta+2\right)$$
Introducing $x=\cos \theta$, we can re-write the above results as
\left{\begin{aligned} A_{w s} & =\pi R^2\left(1-x^2\right), \ A_{a w} & =2 \pi R^2(1-x), \ V & =\frac{\pi R^3}{3}\left(x^3-3 x+2\right) . \end{aligned}\right.

The variations of these quantities are then obtained from
\left{\begin{aligned} d A_{w s} & =2 \pi R\left[\frac{d R}{d x}\left(1-x^2\right)-R x\right] d x \ d A_{a w} & =2 \pi R\left[2 \frac{d R}{d x}(1-x)-R\right] d x \ d V & =\pi R^2\left[\frac{d R}{d x}\left(x^3-3 x+2\right)+R\left(x^2-x\right)\right] d x=0 . \end{aligned}\right.
From the last equation, we conclude
$$\frac{1}{R} \frac{d R}{d x}=-\frac{x^2-x}{x^3-3 x+2}=-\frac{x+1}{(x-1)(x+2)}$$

Substituting for $d R / d x$ gives,
$$d A_{w s}=2 \pi R^2 \frac{d x}{x+2}, \quad \text { and } \quad d A_{a w}=2 \pi R^2 \frac{x \cdot d x}{x+2},$$
resulting in the required result of
$$d A_{a w}=x \cdot d A_{w s}=d A_{w s} \cos \theta .$$
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water-gas interface, (2) the surface tension at the solid-water interface, (3) the surface tension at the gas-solid interface, and (4) the force downward by solid-contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the $y$-direction $x$-directions. The latter gives the condition for the contact angle known as Young’s equation,
$$\mathcal{S}{a s}=\mathcal{S}{a w} \cdot \cos \theta+\mathcal{S}{w s}, \Longrightarrow \cos \theta=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}} .$$
The critical condition for the complete wetting occurs when $\theta=0$, or $\cos \theta=1$, i.e. for
$$\cos \theta_C=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}}=1$$ Complete wetting of the substrate thus occurs whenever $$\mathcal{S}{a w} \leq \mathcal{S}{a s}-\mathcal{S}{w s} .$$

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

• Typical length scales at which the surface tension effects become significant are given by the condition that the forces exerted by surface tension and relevant pressures become comparable, or by the condition that the surface energy is comparable to the other energy changes of interest.

Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy,
$$S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g$$

$$R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} .$$
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about $1 \mathrm{~atm})=$ surface tension of water, gives
$$\pi_{g e l} \approx \frac{N}{V} k_B T \approx \frac{2 S}{R}$$
where $N$ is the number of counter ions within the gel. Thus,
$$R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} .$$