Common Test Physics 2024

Key idea: right at the tipping point, the only torques about the pivot corner are from the push $F$ and from gravity.

Put the pivot at the corner touching the wall. As $F$ grows, the plate tries to swing about that corner, and the other bottom corner lifts off the floor. At the moment it is about to lift, the floor’s push is concentrated at the pivot, so neither the floor force nor the wall force has any torque about it.

• The push $F$ is horizontal and acts at the top corner, a height $L$ above the floor. Its lever arm about the pivot is $L$, so its torque is $FL$.
• Gravity $Mg$ acts at the center of mass. That point is $L/3$ from the vertical edge, so it is $L - L/3 = 2L/3$ from the pivot horizontally. Its torque is $Mg\cdot\tfrac{2L}{3}$, turning the other way.

The plate stays put as long as gravity’s torque wins:

$$FL \le \tfrac{2}{3}MgL \quad\Longrightarrow\quad F_{\max} = \tfrac{2}{3}Mg.$$

That is choice 5.

Key idea: the mean kinetic energy of one particle in a monatomic ideal gas is $\tfrac32 k_B T$. Mass does not appear.

Box 2. Compare a helium nucleus in the core with a helium atom in room-temperature air. The ratio of energies is just the ratio of temperatures: $$\frac{1.5\times10^{7}\ \text{K}}{300\ \text{K}} = 5\times10^{4} = 50000.$$ That is choice 5.

Box 3. Hydrogen and helium nuclei in the core share the same temperature. Since $\tfrac32 k_B T$ does not care about mass, their average kinetic energies are equal: the ratio is $1$, choice 3. (The lighter hydrogen nuclei move faster, but they carry the same energy on average.)

Key idea: across parallel layers, $n\sin\theta$ stays the same, so the glass drops out of the condition.

Snell’s law at each flat boundary keeps the quantity $n\sin\theta$ the same: $$n\sin\theta = n'\sin\theta' = n''\sin\theta''.$$

Where can light be trapped? Going from water into glass, the index rises ($n' > n$). Light bends toward the normal and can always get through, so there is no total reflection there. Going from glass into air, the index falls, so that is the only boundary where total reflection can happen.

The critical angle. Light leaves into air only if $\sin\theta'' \le 1$. With $n'' = 1$, that means $n\sin\theta \le 1$. The light stops coming out when $$\sin\theta_C = \frac{1}{n}.$$

The glass index $n'$ cancels out. The glass layer changes the path, but not the critical angle measured in the water. So the answer is “glass and air” with $1/n$: choice 4.

Key idea: the force $q\vec v\times\vec B$ is perpendicular to $\vec B$, and it vanishes when $\vec v \parallel \vec B$.

Circular motion in the $xy$ plane. The velocity and the force (which points to the center) both lie in the $xy$ plane. The force is perpendicular to $\vec B$. For a uniform field, the circle lies in the plane perpendicular to $\vec B$, so $\vec B$ must point along the $z$ axis.

Straight-line motion along $x$. With gravity ignored, the particle goes straight only if there is no force at all. The magnetic force is zero exactly when the velocity is parallel to the field. So $\vec B$ is along the $x$ axis.

The pair ($z$, $x$) is choice 7.

Key idea: if mass goes down, the missing mass comes out as energy. And $1/16 = (1/2)^4$.

Energy. Before the reaction, the proton and the carbon-12 nucleus total $$1.0073 + 11.9967 = 13.0040\ \text{u}.$$ The nitrogen-13 nucleus has 13.0019 u. The mass dropped by 0.0021 u, so energy was released, about $0.0021\times 931\ \text{MeV} \approx 2\ \text{MeV}$. This does not depend on how fast the proton was going. The proton’s kinetic energy only adds to whatever is released.

Half-life. After $k$ half-lives, the fraction left is $(1/2)^k$. One-sixteenth is $(1/2)^4$, so 40 minutes is 4 half-lives, and the half-life is 10 minutes.

“Released” with 10 minutes is choice 7.

Key idea: water is incompressible, so the volume that leaves the nozzle equals the volume by which the water inside drops.

In a short time $\Delta t$, the water that leaves forms a cylinder with cross-section $s$ and length $u\,\Delta t$: $$\Delta V = s\,u\,\Delta t.$$ In the same time, the surface inside the wide bottle falls by $u_0\,\Delta t$, which uncovers a volume $S_0 u_0\,\Delta t$. Setting these equal gives $$u_0 = \frac{s}{S_0}\,u.$$ A narrow nozzle (small $s$) makes the inside surface creep down slowly. That is choice 6.

Key idea: mass is density times volume, and a gas at steady pressure does work $p\,\Delta V$.

• Mass. The water that leaves has volume $\Delta V$ and density $\rho_0$, so $\Delta m = \rho_0\,\Delta V$ (choice 2).
• Work. As the water leaves, the compressed air expands into the space it left behind, an extra volume $\Delta V$. Over the short interval its pressure is about constant at $p$, so the air does work $W' = p\,\Delta V$ (choice 1). The work is positive because the gas expands.

Key idea: with no friction, the air’s work all turns into the kinetic energy of the water jet.

Set the kinetic energy of the escaping water equal to the work done by the air: $$\tfrac12\,\Delta m\,u^2 = W' \quad\Longrightarrow\quad u = \sqrt{\frac{2W'}{\Delta m}}.$$ So the blanks are “kinetic energy” and $\sqrt{2W'/\Delta m}$, choice 9.

With the last part, $W' = p\,\Delta V$ and $\Delta m = \rho_0\,\Delta V$, so $$u = \sqrt{\frac{2p}{\rho_0}}.$$ This is the same speed Bernoulli’s equation gives for a jet driven by pressure $p$.

Key idea: the total momentum of rocket plus ejected water stays at zero.

Before release, everything is at rest, so the total momentum is $0$. With no outside forces in the model (gravity is ignored for this step), the total stays $0$ after $\Delta t$.

Take upward as positive. The rocket (mass about $M$) moves up with speed $\Delta v$, so its momentum is $+M\,\Delta v$. The water slug (mass $\Delta m$) shoots down at speed $u$, so its momentum is $-\Delta m\,u$. Then $$M\,\Delta v - \Delta m\,u = 0.$$ That is choice 4. Notice that momentum, not kinetic energy, is what stays the same here. The air’s push turns stored energy into motion, so kinetic energy is not conserved.

Key idea: the thrust is the rate at which the rocket gains momentum, $M\,\Delta v/\Delta t$.

The water pushes on the rocket with an average force equal to the rocket’s momentum change per unit time: $$F_\text{thrust} = \frac{M\,\Delta v}{\Delta t}.$$ The rocket rises only if this beats its weight: $$\frac{M\,\Delta v}{\Delta t} > Mg \quad\Longleftrightarrow\quad \Delta v > g\,\Delta t.$$ That is choice 4. It says the rocket must gain speed faster than free fall would take it away.

A nice check: by the last part, $M\,\Delta v = \Delta m\,u$, so $F_\text{thrust} = \rho_0 s u^2 = 2ps$. The thrust is twice the pressure times the nozzle area.

Key idea: $\vec F = I\vec L\times\vec B$, so the force is perpendicular to both the current and the field.

The current runs along the wire ($x$ direction). The magnet’s field points along $y$. The force is perpendicular to both, so it is along the $z$ axis, and it flips back and forth as the AC current reverses.

This push acts at the middle of the string. A spot that is shaken hard cannot be a node, which never moves. It is an antinode.

($z$ axis, antinode) is choice 5.

Key idea: one loop is $\lambda/2$.

Both ends of the string are fixed, so they are nodes. A standing wave with $n$ antinodes has $n$ loops between the ends, and each loop is half a wavelength: $$n\cdot\frac{\lambda}{2} = L \quad\Longrightarrow\quad \lambda = \frac{2L}{n}.$$ For three loops, $\lambda = 2L/3$, choice 3.

Key idea: $f_n = v/\lambda_n = n\,\dfrac{v}{2L}$, a straight line through the origin.

With $\lambda_n = 2L/n$ and $v = f\lambda$, $$f_n = \frac{v}{\lambda_n} = \frac{v}{2L}\,n.$$ Plotting $f_n$ against $n$ gives a line through the origin with slope $v/(2L)$. Since $L$ is fixed, the slope is proportional to the wave speed $v$ on the string. That is choice 2. The phase, the amplitude and the current do not set the frequencies.

Key idea: if $f_3 \propto X$, then the plot of $f_3$ against $X$ is a straight line through the origin.

Go through the four graphs:

• Against $S$, the points curve and the line through them misses the origin.
• Against $1/S$, $f_3$ falls as the variable grows, so it cannot be proportional to it.
• Against $S^2$, the points bend over.
• Against $\sqrt S$, the points sit on a straight line that heads toward the origin.

So $f_3 \propto \sqrt S$, choice 2. This matches the theory, $v = \sqrt{S/\mu}$, since $f$ is proportional to $v$.

Key idea: check whether $d\cdot f$ stays about the same.

Multiply each frequency by its diameter (in mm):

Each row is about constant, so $f_n \propto 1/d$, choice 4.

Why. For a wire of one material, the mass per length is $\mu \propto d^2$ (the cross-section area). So $v = \sqrt{S/\mu} \propto 1/d$, and $f$ follows $v$. Together with the last part, this is the experimental route to $v = \sqrt{S/\mu}$.

Key idea: $V = kq\left(\dfrac1{r_+} - \dfrac1{r_-}\right)$. It is zero on the perpendicular bisector and grows steeply near each charge.

Three features pick out the right picture:

1. Near each charge the potential behaves like a single point charge, so the equipotentials are small closed loops around it. They are crowded, because the potential changes fast there.
2. On the perpendicular bisector, each point is the same distance from both charges, so the two terms cancel and $V = 0$. That bisector is a single straight equipotential line.
3. Farther from the charges, the field gets weaker, so lines drawn at equal steps of potential spread apart. The curves wrap around each charge and get wider and flatter as they approach the middle line.

The picture with loops around each charge, one straight line down the middle, and widening gaps is choice 2. Straight parallel lines would mean a uniform field, which two point charges cannot make.

Key idea: field lines show strength by their density and cross equipotentials at right angles. Equal spacing only happens in a uniform field.

• Field lines are denser where the field is stronger. True. That is how field-line diagrams show strength.
• Neighboring equipotentials are always the same distance apart. False. For equal steps $\Delta V$, the spacing is about $\Delta V/E$, so it shrinks where the field is strong. Around the dipole, the lines bunch up near the charges.
• Field lines and equipotentials meet at right angles. True. Moving along an equipotential takes no work, so the field has no component along it.

The first and third are true: choice 5.

Key idea: the field crosses equipotentials at right angles, and the current density follows the field, $\vec J = \vec E/\rho$.

Near an edge, the measured equipotentials meet the edge at right angles. The field is perpendicular to the equipotentials, so near the edge the field runs parallel to the edge.

In an ordinary conductor, Ohm’s law says the current flows in the same direction as the field. So near the edge the current also runs parallel to the edge. It has to: charge cannot flow out through the edge of the paper.

(parallel, same, parallel) is choice 1.

Key idea: $E = \left|\dfrac{\Delta V}{\Delta x}\right|$, the slope of the $V$–$x$ graph.

Near $x = 0$, the points rise by about $0.1$ mV for every $15$ mm. For example, the potential goes from 0 mV at $x=0$ to 0.1 mV near $x = 15$ mm. So $$E \approx \frac{0.1\times10^{-3}\ \text{V}}{15\times10^{-3}\ \text{m}} \approx 6.7\times10^{-3}\ \text{V/m}.$$ The closest choice is $7\times10^{-3}$ V/m, choice 6.

Don’t drop the units: mV over mm is $10^{-3}/10^{-3}$, so the slope in mV/mm is the same number in V/m.

Key idea: for a short piece of length $\Delta x$, the voltage is $E\,\Delta x$ and the resistance is $\rho\,\Delta x/S$.

Take a thin slab of the paper of thickness $\Delta x$ around $x = 0$, with cross-section $S$.

• The voltage across it is $\Delta V = E\,\Delta x$.
• Its resistance is $R = \rho\,\dfrac{\Delta x}{S}$.
• Ohm’s law, $\Delta V = IR$, gives $E\,\Delta x = I\rho\,\dfrac{\Delta x}{S}$.

The $\Delta x$ cancels: $$\rho = \frac{SE}{I}.$$ That is choice 1. It is the same as the local form of Ohm’s law, $J = I/S = E/\rho$.