Assuming the atoms have the same mass, a collision between atoms in a vacuum is most nearly an elastic collision. This means the kinetic energy of the system is conserved and when they collide, all of the kinetic energy in atom A will be transferred perfectly to atom B. So just after collision, atom A will have velocity $\displaystyle 0$ and atom B will have velocity $\displaystyle v$. Note that momentum is always conserved.

The equation for conservation of momentum is:

$\displaystyle m_{i}v_{i}=m_{f}v_{f}$

Given:

$\displaystyle m_{i}=3000kg$

$\displaystyle v_{i}=40\frac{km}{hr}$

$\displaystyle m_{f}=3000kg_{truck}+1800kg_{sand}$

Plug in given values into the conservation of momentum equation.

$\displaystyle 3000kg\cdot 40\frac{km}{hr}=4800kg\cdot v_{f}$

Simplify.

$\displaystyle 120000kg\cdot 1\frac{km}{hr}=4800kg\cdot v_{f}\\$

$\displaystyle v_{f}=\frac{1200}{48}\frac{km}{hr}=\frac{100}{4}\frac{km}{hr}=25\frac{km}{hr}$

 $\displaystyle v_f=25\frac{km}{hr}$

Subtract the final velocity from the initial velocity to find change in speed (since these velocities are in the same direction).

 $\displaystyle 40\frac{km}{hr}-25\frac{km}{hr}=15\frac{km}{hr}$.

Because of conservation of momentum, the product of the initial mass and velocity should equal the final mass and velocity. The equation for conservation of momentum is:

$\displaystyle m_{i}v_{i}=m_{f}v_{f}$

The initial mass of the truck is $\displaystyle 3000kg$

The initial velocty is $\displaystyle 40\frac{km}{hr}$

The final velocity is $\displaystyle 4km/hr$ 

Plug in known values to the conservation of momentum equation.

$\displaystyle 3000kg\cdot 40\frac {km}{hr}=m_{f}\cdot 4\frac{km}{hr}$

Simplify.

$\displaystyle 3000kg\cdot 10=m_{f}$

$\displaystyle 30000kg=m_{f}$

Note that while necessary, the final mass is not our final answer. The final mass, $\displaystyle m_{f}$, is the mass of the truck, $\displaystyle m_{truck}$, and the mass of the sand added to the truck, $\displaystyle m_{sand}$.

The mass of the truck is known so set up an equation and solve for the mass of sand added to the truck.

$\displaystyle m_{f}=m_{truck}+m_{sand}$

$\displaystyle 30000kg = 3000kg-m_{sand}$

$\displaystyle m_{sand}=27 000kg$

Relevant equations:

$\displaystyle E = \frac{kq}{r^2}$ (electric field of point charge)

Anywhere outside the metal sphere, the electric field is the same as it would be for a point charge of the same magnitude, located at the center of the sphere. So, calculate the electric field of a point charge given:

$\displaystyle q = 4.0 * 10^{-6}C$

$\displaystyle r = 2.0m+0.05m=2.05m$

$\displaystyle k = 9.0 * 10^9 \frac{N\cdot m^2}{C^2}$

Plugging in gives:

$\displaystyle E = \frac{(9.0*10^9\frac{N\cdot m^2}{C^2})(4.0 * 10^{-6}C)}{(2.05m)^2}=\frac{36000}{4.2} \frac{N}{C}=8.6*10^3 \frac{N}{C}$

Relevant equations:

$\displaystyle E = \frac{\sigma }{2 \epsilon _o}$  (field due to single infinite plane)

Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields:

$\displaystyle E_{total} = E_1 + E_2$

$\displaystyle E _{total} = \frac{\sigma}{2 \epsilon _o}+\frac{\sigma}{2 \epsilon _o}$

$\displaystyle E_{total}=\frac{\sigma}{\epsilon_o}$

The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.

The correct answer is 45 degrees upwards of left. Since all particles have charge $\displaystyle +q$, all forces will be repulsive (there will be no attracting forces). The particle in the top-right corner creates a repulsive force directly to the left, and the particle in the bottom-left corner creates a repulsive force directly upwards. These are equal in magnitude, since they are both at distance $\displaystyle r$ from the top left corner. The bottom-right corner also creates a repulsive force, but acts along the same direction as the vector sum of a leftwards and upwards force.

To solve this problem, we will need to derive an equation.

We know that:

$\displaystyle C=\left | \frac{Q}{V}\right |$

$\displaystyle V=\oint E\cdot dr$

We can use Gauss's law to derive the electric field between the two circles yielding:

 $\displaystyle E = \frac{Q}{\epsilon_{0}4\pi r^2}$

Doing our integration with respect to $\displaystyle dr$ from $\displaystyle R_1$ to $\displaystyle R_2$, we get:

$\displaystyle V = \frac{-Q(R_{2} - R_{1})}{\epsilon _{0}4\pi R_{1}R_{2}}$

We can plug this back into our equation for capacitance to get:

$\displaystyle C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

Coulomb's law for the electric field from point charges is $\displaystyle E=\frac{kq}{r^2}$, where we know the values of the following variables.

$\displaystyle k=8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2}$

$\displaystyle q=100\ \text{nC}= 100*10^{-9}\ \text{C}$

$\displaystyle r=500\ \mu\text{m}= 500*10^{-6}\ \text{m}$

Using these values, we can solve for the electric field.

$\displaystyle E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(100*10^{-9}\ \text{C})}{(500*10^{-6}\ \text{m})^2}$

$\displaystyle E=3.6* 10^9 \frac{\text{N}}{\text{C}}$

To find the location at which the test charge experience zero net force, write the net force equation as $\displaystyle F_{net}=F_1-F_2=0$, where $\displaystyle F_1$ is the force on the test charge from $\displaystyle q_1$, and $\displaystyle F_2$ is the force on the same test charge from $\displaystyle q_2$. Using Coulomb's law, we can rewrite the force equation and set it equal to zero.

$\displaystyle F=\frac{kq_1q_2}{r^2}$

$\displaystyle F_{net}=\frac{kq_1q}{r^2}-\frac{kq_2q}{(d-r)^2}=0$

In this equation, the distance, $\displaystyle r$, is how far away the test charge is from $\displaystyle q_1$, while $\displaystyle (d-r)$ represents how far away the test charge is from $\displaystyle q_2$.  Now, we simplify and solve for $\displaystyle r$.

Cross-multiply.

$\displaystyle kq_1q(d-r)^2=kq_2qr^2$

We can cancel $\displaystyle k$ and $\displaystyle q$. We do not need to know these values in order to solve the question.

$\displaystyle q_1(d-r)^2=q_2r^2$

$\displaystyle \sqrt{q_1}(d-r)=\sqrt{q_2}r$

$\displaystyle \sqrt{q_1}d-\sqrt{q_1}r=\sqrt{q_2}r$

$\displaystyle \sqrt{q_1}d=r(\sqrt{q_2}+\sqrt{q_1})$

$\displaystyle r=\frac{\sqrt{q_1}d}{\sqrt{q_2}+\sqrt{q_1}}$

Now that we have isolated $\displaystyle r$, we can plug in the values given in the question and solve.

$\displaystyle r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(1000*10^{-6}\ \text{m})}{\sqrt{32*10^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$\displaystyle r=\frac{(1.45*10^{-4}\ \text{C})(1000*10^{-6}\ \text{m})}{1.79*10^{-4}\ \text{C}+1.45*10^{-4}\ \text{C}}$

$\displaystyle r=4.5*10^{-4}\ \text{m}$

Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, $\displaystyle mg$. In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, $\displaystyle qE$. To show that you would float, the net force equation is written as $\displaystyle F_{net}=qE-mg=0$, where $\displaystyle q$ is the charge on you.

For plates that are charged, know that $\displaystyle E=\frac{\sigma}{2\epsilon_0}$.

Knowing this, the force equation becomes $\displaystyle q\frac{\sigma}{2\epsilon_0}-mg=0$.

Solve for $\displaystyle q$.

$\displaystyle q=\frac{2\epsilon_0mg}{\sigma}$

Now we can plug in our given values, and solve for the charge.

$\displaystyle q=\frac{2(8.85*10^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{9*10^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$\displaystyle q=\frac{1.2*10^{-8}\frac{C^2}{m^2}}{9*10^{-8}\frac{C}{m^2}}$

$\displaystyle q=0.13\ \text{C}$