Structured Questions

Question 1

Electric charge is distributed uniformly along an infinitely long thin wire. The charge per unit length is $\lambda$. Two points A and B are located along the radial line perpendicular to the central axis of the wire.

a. Using Gauss’s Law, determine the expression of the electric field $E_r$ at a distance $r$ from the wire.

b. In the figure, it is given that the charge per unit length is $\lambda =2.5\times {10}^{-6}\text{C/m}$. A small object of mass $m=0.002\text{kg}$ with charge $q=8.5\mu \text{C}$ is initially at rest at point A.

i. Determine the initial acceleration (in m/s${}^2$) of the small object at point A, $r_A=0.030\text{m}$ from the wire when it is released.

ii. Given that the electrical potential difference between points A and B is given by

determine the speed (in m/s) of the object when it reaches point B where $r_B=0.050\text{m}$.

Solution

Part a. Electric Field using Gauss's Law

The goal here is to find the electric field $\overrightarrow{E}$ created by the infinitely long wire at a distance $r$ from its axis.

1. Identify the Symmetry and Choose a Gaussian Surface

Since the wire is infinitely long and has a uniform linear charge density $\lambda$, the electric field $\overrightarrow{E}$ must:

• Be radially outward (if $\lambda$ is positive) or radially inward (if $\lambda$ is negative) from the wire. It cannot have a component parallel to the wire due to the wire's infinite length (symmetry).
• Have the same magnitude at all points equidistant from the wire.

The ideal Gaussian surface that exploits this symmetry is a coaxial cylinder of radius $r$ and arbitrary length $L$, centered on the wire.

2. Apply Gauss's Law

Gauss's Law states that the total electric flux (${\mathrm{\Phi }}_E$) through a closed surface (the Gaussian surface) is proportional to the total electric charge ($q_{enc}$) enclosed within that surface:

where ${\epsilon }_0$ is the permittivity of free space.

3. Calculate the Electric Flux (${\mathrm{\Phi }}_E$)

The cylindrical Gaussian surface has three parts: two flat circular ends and the curved side.

• Ends: The electric field $\overrightarrow{E}$ is perpendicular to the ends' area vectors $d\overrightarrow{A}$. Thus, $\overrightarrow{E}\cdot d\overrightarrow{A}=0$, and the flux through the ends is zero.
• Curved Side: On the curved side, $\overrightarrow{E}$ is perpendicular to the surface (radial) and has constant magnitude $E_r$ at a fixed radius $r$. The area vector $d\overrightarrow{A}$ is also perpendicular to the surface (radial). Thus, $\overrightarrow{E}$ and $d\overrightarrow{A}$ are parallel, so $\overrightarrow{E}\cdot d\overrightarrow{A}=E_{r}dA$.

The total flux is:

The integral ${\int }_{\text{curved}}dA$ is the area of the curved surface of the cylinder, which is $2\pi rL$.

4. Calculate the Enclosed Charge ($q_{enc}$)

The charge enclosed within the Gaussian cylinder of length $L$ is determined by the linear charge density $\lambda$:

5. Solve for the Electric Field ($E_r$)

Substitute ${\mathrm{\Phi }}_E$ and $q_{enc}$ back into Gauss's Law:

The length $L$ cancels out, which is expected since the result shouldn't depend on the arbitrary length of our Gaussian surface.

The expression for the electric field $E_r$ at a distance $r$ from the wire is:

Part b. Dynamics and Energy

Now we'll use this result, along with the given values, to analyze the motion of a charged object.

Given values:

• Linear charge density, $\lambda =2.5\times {10}^{-6}\text{C/m}$
• Object mass, $m=0.002\text{kg}$
• Object charge, $q=8.5\mu \text{C}=8.5\times {10}^{-6}\text{C}$
• Permittivity of free space, ${\epsilon }_0\approx 8.854\times {10}^{-12}{\text{C}}^2/(\text{N}\cdot {\text{m}}^2)$

i. Initial Acceleration at Point A

To find the acceleration, we first need to find the electric force on the object using Newton's Second Law ($F=ma$) and the relationship between Electric Force and Field ($F=qE$).

1. Calculate the Electric Field at $r_A$

Using the result from part a with $r_A=0.030\text{m}$:

We can use the constant $k=\frac{1}{4\pi {\epsilon }_0}\approx 9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$. Note that $\frac{1}{2\pi {\epsilon }_0}=2\times \frac{1}{4\pi {\epsilon }_0}=2k$.

2. Calculate the Electric Force at $r_A$

The force $\overrightarrow{F}$ is in the direction of the field $\overrightarrow{E}$ since $q$ is positive.

3. Calculate the Initial Acceleration ($a_A$)

Using Newton's Second Law, $F=ma$:

The initial acceleration of the object is $\mathbf{6375}{\text{m/s}}^{\mathbf{2}}$, directed radially away from the wire.

ii. Speed at Point B

To find the speed at point B, where $r_B=0.050\text{m}$, we can use the Conservation of Energy. The only non-conservative force acting is the electric force, so the work done by the electric force equals the negative change in potential energy, which is also related to the potential difference.

1. Relate Work, Potential Difference, and Kinetic Energy

The work done by the electric field as the charge moves from A to B ($W_{AB}$) is:

The potential difference $V_{AB}$ is defined as $V_A-V_B$.

By the Work-Energy Theorem, this work must equal the change in kinetic energy:

Since the object starts at rest at A, the initial kinetic energy $K_A=0$.

Therefore:

2. Substitute the Expression for $V_{AB}$

We are given:

Substitute this into the energy equation:

3. Calculate $V_{AB}$

First, let's calculate the numerical value for $V_{AB}$:

• Ratio: $\frac{r_B}{r_A}=\frac{0.050\text{m}}{0.030\text{m}}=\frac{5}{3}\approx 1.6667$
• $2k\lambda =2(9.0\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2)(2.5\times {10}^{-6}\text{C/m})=4.5\times {10}^4\text{N}\cdot \text{m}/\text{C}$ (or V/m)

4. Solve for $v_B$

Rearrange the energy equation to solve for $v_B$:

Substitute the values:

The speed of the object when it reaches point B is approximately $\mathbf{13.98}\text{m/s}$.

Question 2

Cycling events in 2020 Tokyo Olympics featured five disciplines – BMX Freestyle, BMX Racing, Road, Mountain and Track Cycling. Track cycling takes place in a velodrome where the sides are banked. This allows track cyclists to reach maximum speeds that are far greater than the average speed when cycling on a flat road. Fig. 2.2 and 2.3 shows an object rounding a curve on a flat surface and on a banked surface in a velodrome respectively. The radius of the path that the object makes, $r=20\text{m}$. The coefficient of static friction of the surfaces, ${\mu }_s=0.70$. You may assume the object has negligible thickness.

a. Referring to Fig 2.2, determine the maximum speed (in m/s) at which the object can safely travel around the flat curve without skidding.

b. At maximum speed $v_{max}$ in the velodrome, friction on the object acts down the bank as shown in Fig. 2.3.

i. In addition to the frictional force identified for you, sketch labelled arrows to show the other two other forces acting on the object.

ii. Apply Newton’s second law on the object in the vertical and horizontal directions.
Vertical:

Horizontal:

iii. Using your answers in ii or otherwise, show that the maximum speed of the object

iv. If the stadium is built with angle $\theta ={44}^{\circ }$ with the bank above the horizontal, determine the maximum speed (in m/s) of the cyclist.

Solutions

Part a. Flat Curve (Fig. 2.2)

For an object to safely travel around a flat, horizontal curve without skidding, the force that provides the necessary centripetal acceleration ($\frac{m{v}^2}{r}$) must come entirely from the static friction force ($f_s$). The maximum speed occurs when the static friction reaches its maximum possible value, $f_{s,\text{max}}={\mu }_{s}N$.

1. Vertical Analysis (Forces Perpendicular to Motion):

   On a flat road, the forces in the vertical direction (perpendicular to the surface) are the normal force ($N$) and the gravitational force ($mg$). Since there is no vertical acceleration, they must balance:

1. Horizontal Analysis (Forces Providing Centripetal Force):
   The maximum static friction provides the required centripetal force ($F_c$):

1. Solve for $v_{\text{max}}$:

Substitute $N=mg$ into the horizontal equation:

The mass ($m$) cancels out:

1. Calculation:

Using the given values: $r=20\text{m}$, ${\mu }_s=0.70$, and $g\approx 9.8{\text{m/s}}^2$:

The maximum speed the object can safely travel around the flat curve is $\mathbf{11.7}\text{m/s}$.

Part b. Banked Curve with Friction (Fig. 2.3)

i. Sketch of Forces

At the maximum speed $v_{max}$, the object is trying to slide up the bank, so the static friction force ($f$) acts down the bank (as shown in Fig. 2.3). The other two forces acting on the object are:

1. Gravitational Force ($mg$): Acting straight downward.
2. Normal Force ($N$): Acting perpendicular to the banked surface (up and away from the bank).

ii. Application of Newton’s Second Law

We will set up a coordinate system where the horizontal direction is towards the center of the circle (radial direction) and the vertical direction is straight up.

1. Resolve Forces into Components

The angle of the bank is $\theta$.

• Normal Force ($N$):
  • Horizontal component: $N\sin \theta$ (towards the center)
  • Vertical component: $N\cos \theta$ (upward)
• Friction Force ($f$): Since friction is parallel to the bank and down the bank:
  • Horizontal component: $f\cos \theta$ (towards the center)
  • Vertical component: $f\sin \theta$ (downward)
• Gravitational Force ($mg$):
  • Horizontal component: $0$
  • Vertical component: $mg$ (downward)

2. Vertical Equilibrium ($\mathrm{\Sigma }{F}_y=0$)

The object does not accelerate vertically.

At maximum speed $v_{max}$, the friction is maximum static friction: $f=f_{s,\text{max}}={\mu }_{s}N$. We'll use $\mu$ for ${\mu }_s$ for simplicity in the derivation.

This matches the given vertical equation result.

3. Horizontal Motion ($\mathrm{\Sigma }{F}_x=F_c$)

The net horizontal force provides the centripetal force $F_c=\frac{m{v}_{max}^2}{r}$. All horizontal components point towards the center (positive radial direction).

Again, using $f=\mu N$:

This matches the given horizontal equation result.

iii. Derivation of $v_{max}$

We have two independent expressions for the Normal Force $N$. We can set them equal to each other to solve for $v_{max}$:

1. Cancel mass ($m$):

1. Solve for $v_{max}^2$:

1. Take the square root:

The derivation is complete and matches the required expression. This equation clearly shows that banking the curve ($\theta >0$) increases $v_{max}$ significantly compared to a flat curve ($\theta =0$, where $v_{max}=\sqrt{rg\mu }$).

iv. Maximum Speed Calculation

We use the derived formula with the given values:

• $r=20\text{m}$
• $\mu =0.70$
• $\theta ={44}^{\circ }$
• $g=9.8{\text{m/s}}^2$

First, calculate the trigonometric values:

• $\sin {44}^{\circ }\approx 0.6947$
• $\cos {44}^{\circ }\approx 0.7193$

Numerator:

$rg(\mu \cos \theta +\sin \theta )$

$=(20)(9.8)[(0.70)(0.7193)+0.6947]$

$=196[0.5035+0.6947]$

$=196(1.1982)\approx 234.847$

Denominator:

$\cos \theta -\mu \sin \theta$

$=0.7193-(0.70)(0.6947)$

$=0.7193-0.4863$

$\approx 0.2330$

Calculate $v_{max}$:

$v_{max}=\sqrt{\frac{234.847}{0.2330}}$

$v_{max}=\sqrt{1007.927{\text{m}}^2/{\text{s}}^2}$

${\mathbf{v}}_{max}\approx \mathbf{31.7}\text{m/s}$

The maximum speed of the cyclist on the banked track is $\mathbf{31.7}\text{m/s}$. That's quite a bit faster than the $11.7\text{m/s}$ on the flat road!

Question 3

In the following cases, the mass of the bullet $m_1=10\text{g}$, and the mass of the block $m_2=0.30\text{kg}$. A bullet is fired at speed $u_1$ into a block. After hitting and penetrating the block, the bullet and the block move at a common velocity $v$ on a horizontal frictionless surface and is observed to compress the spring by $x=0.30\text{m}$.

i. In terms of $u_1$, determine the expression for the common speed $v$ at which the block and bullet move.

ii. If the spring constant of the spring is $k=720\text{N/m}$, determine the speed (in m/s) of the bullet $u_1$.

iii. In reality, the surface has friction. If all the given values remain the same, how will the actual speed of the bullet be as compared to the value of $u_1$ calculated in (ii)?

A. Larger
B. The same
C. Smaller

Solutions

Here are the given values:

• Mass of bullet, $m_1=10\text{g}=0.010\text{kg}$
• Mass of block, $m_2=0.30\text{kg}$
• Initial speed of bullet, $u_1$ (to be found)
• Common final speed after collision, $v$
• Maximum compression of spring, $x=0.30\text{m}$
• Spring constant, $k=720\text{N/m}$
• Surface is frictionless for parts (i) and (ii).

i. Expression for Common Speed $v$

The collision between the bullet and the block is a perfect inelastic collision because the two objects stick together and move with a common final velocity $v$. In the absence of external forces (like friction on the whole system), Linear Momentum is conserved during the collision.

Initial Momentum ($P_{\text{initial}}$):

Since the block is initially at rest, $u_2=0$.

Final Momentum ($P_{\text{final}}$):

After the collision, the bullet and block move together with speed $v$.

Conservation of Momentum ($P_{\text{initial}}=P_{\text{final}}$):

Solving for the common speed $v$:

Substituting the numerical masses:

The expression for the common speed $v$ in terms of $u_1$ is:

ii. Determine the Speed of the Bullet $u_1$

After the collision, the combined mass $(m_1+m_2)$ moves toward the spring. As the mass compresses the spring, the Kinetic Energy of the mass is converted into Elastic Potential Energy stored in the spring. Since the surface is frictionless, Mechanical Energy is conserved during this process.

Initial Energy (before spring compression):

The energy is purely kinetic energy of the combined mass:

Final Energy (at maximum compression $x$):

At maximum compression, the mass is momentarily at rest, so $K_{\text{final}}=0$. The energy is purely elastic potential energy:

Conservation of Energy ($E_{\text{initial}}=E_{\text{final}}$):

Step 1: Solve for the common speed $v$

Substituting the numerical values:

Step 2: Use the result from (i) to find $u_1$

From part (i): $v=\frac{m_1}{m_1+m_2}{u}_1=\frac{1}{31}{u}_1$

The initial speed of the bullet $u_1$ is approximately $448\text{m/s}$ (to 3 significant figures).

iii. Effect of Friction

In this hypothetical scenario, the surface has friction, but all other values ($m_1,m_2,k,x$) remain the same.

The total mechanical energy loss due to friction is equal to the work done by friction, $W_{\text{friction}}$. This work is negative because the friction force opposes the motion.

Energy Conservation with Friction:

Since $W_{\text{friction}}$ is negative, the initial kinetic energy $\frac{1}{2}(m_1+m_2)v_{\text{actual}}^2$ must be greater than the stored potential energy $\frac{1}{2}k{x}^2$ by the amount of energy lost to friction.

Because $-W_{\text{friction}}$ is a positive value (energy added to the right side of the equation):

This means the actual common speed $v_{\text{actual}}$ just after the collision must be larger than the speed $v$ calculated in part (ii) in the frictionless case to achieve the same compression $x$.

Since $u_1$ is directly proportional to $v$ ($u_1\propto v$):

If $v_{\text{actual}}$ is larger, the initial speed of the bullet $u_{1,\text{actual}}$ must also be larger than the value calculated in part (ii) to compensate for the energy loss due to friction.

The correct choice is A. Larger.