Physics Error Book

A running error book for physics. Every question I have answered wrongly is filed here under its chapter, with the working that repairs it and my original mistake kept one click away — options can be clicked, key answers checked, and any card cleared and tried again.

Force and Motion

01Multiple choiceEnergy conservationWork done against frictionInclined plane

A block is projected upwards along a rough incline which has a constant frictional force acting on the block.

Figure 1. A block projected up a rough incline, having travelled a distance s along the slope

Which of the following correctly shows the variation of potential energy

and kinetic energy

with distance

that the block travelled up the incline? Neglect air resistance.

Figure 2. The four candidate graphs of energy against s

A rises as a straight line; falls as a straight line, steeper than rises.
B rises as a straight line; falls along a curve that steepens as increases.
The steepening curve claims the block loses kinetic energy faster the further it travels. That is the “it slows down more and more” intuition — falling speed read as rising deceleration. The retarding force is , fixed by the inclination and by the constant friction the stem gives, so cannot change along the slope.
C rises as a straight line; falls along a curve that flattens as increases.
Correct reasoning applied to the wrong axis. Against time the flattening curve is right — under constant deceleration. Against distance it is not: makes linear in . The horizontal axis here is .
D rises along a curve that flattens; falls along a curve that flattens.
is already wrong before is considered. The height gained is on a straight incline, so is a straight line of gradient whatever friction does. A flattening describes a ramp whose slope decreases with distance, not the constant inclination drawn.

Correct answerA

Solution

Let be the inclination of the incline; be the mass of the block; be the frictional force acting on the block; and and be the initial potential energy and the initial kinetic energy of the block respectively.

The vertical distance that the block travelled up the incline is .

By , the work done against friction by the block is .

Loss in kinetic energy of the block = Gain in potential energy of the block + Work done against friction by the block

Therefore the variation of and with can each be represented by a straight line.

02Multiple choiceNewton's second lawInclined planeProportional reasoning

A block of mass is placed on a smooth incline making an angle with the horizontal as shown. When a force of magnitude parallel to the incline is applied to the block, it travels up the incline with an acceleration . If the applied force becomes , what would the magnitude of the acceleration be ?

Figure 3. A block on a smooth incline of angle theta, with the applied force F and the acceleration a both directed up the slope; below, the four options for the new acceleration

Agreater than
Bequal to
This assumes . Proportionality holds for the net force, not for the applied one, and the net force here is . Doubling doubles the first term and leaves the second untouched, so the net force is more than doubled and so is the acceleration.
Cbetween and
Doubling the acceleration would mean subtracting the weight term twice, because . The block pushed by subtracts it once. The belief behind this option is that gravity keeps eating into the applied force, so the gain must fall short of doubling; the direction is backwards. The retarding term does not grow when grows, so the acceleration grows faster than the force does.
Dwhether it is greater than , equal to or between and depends on the value of
sets the size of the gap, not its sign. The gap is , which is positive for every incline, . A steeper slope puts the answer further above ; no slope puts it below.

I answeredCCorrect answerA

Solution

Let be the mass of the block; be the inclination of the incline; be the acceleration of the block when the applied force is .

The incline is smooth, so the only forces along it are the applied force and the component of the weight. Taking up the incline as positive, Newton’s second law gives

for every inclination, so

Radioactivity

03Multiple choiceNuclear fissionInduced reactionBinding energy

Which statement about nuclear fission is correct ?

AIt involves two nuclei combined together.
That is the definition of nuclear fusion. Fission is one nucleus splitting into two or more nuclei.
BAll nuclear fission reactions are spontaneous.
“Spontaneous” is imported from radioactive decay, which really is spontaneous and random. Fission is normally induced: slow-moving neutrons are needed to initiate the fission of U-235. One counter-example defeats “all”.
CRate of fission reaction is temperature dependent.
Fission is treated here like a chemical reaction, whose rate rises with temperature because the reactants are given more thermal energy. Fission is a nuclear process — the rate is set by the supply of neutrons, not by thermal energy, so it is not temperature dependent.
DFission of heavy nuclei yields products that are more stable in terms of energy.

I answeredBCorrect answerD

Solution

Energy is released in the fission of heavy nuclei, so the products are at a lower energy — that is, more stable — than the original nucleus. In terms of the binding-energy-per-nucleon curve, the fragments lie closer to its peak than the heavy nucleus does.

Electricity and Magnetism

04Multiple choiceTransformer turns ratioPotential dividerOpen circuit

In the circuits below, if a 12 V sinusoidal a.c. is applied across ab and across xy respectively, the voltages across cd and zw are both 6 V. Now if a 6 V sinusoidal a.c. is applied across cd and across zw respectively, what would be the voltages across ab and xy respectively ?

Figure 4. Left: a transformer whose primary is ab and whose tapped secondary gives cd. Right: a potential divider whose full resistor lies across xy, with the slider tapped off to z and w joined to y. Below, the four options for the voltages across ab and xy

AVoltage across ab is 12 V; voltage across xy is 12 V.
The divider is being reversed the way the transformer is — 6 V across the lower section must therefore reappear as 12 V across the whole. A resistor chain has no turns ratio to invert; it can only divide whatever voltage is supplied to it. Nothing is connected to x, so the upper section carries no current and contributes no p.d. to add to the 6 V.
BVoltage across ab is 12 V; voltage across xy is 6 V.
CVoltage across ab is 6 V; voltage across xy is 6 V.
This reads the transformer as a fixed “12 V in, 6 V out” box, or as a device that can only step down. What is fixed is the turns ratio , not either voltage. Interchanging which winding is driven inverts the ratio, so driving cd steps up.
DVoltage across ab is 12 V; voltage across xy is 0 V.
The right premise pushed one step too far. No current flows in the upper section because x is an open circuit — but that makes the p.d. across the upper section zero, not the p.d. across xy. Zero drop from x to z puts x at the potential of z, so xy carries the full 6 V that already appears across zy.

I answeredDCorrect answerB

Solution

Let and be the numbers of turns of the windings ab and cd; and be the resistances of the upper and lower sections of the resistor; and be the current drawn from x.

For the ideal transformer, the ratio of the voltages equals the turns ratio:

Driving cd interchanges the primary and the secondary, and the turns ratio is a property of the windings:

For the potential divider, w and y are the same node, so is the p.d. across alone, and

Nothing is connected to x, so and :

The 2:1 step-down is undone by the transformer but not by the divider because only the transformer’s ratio is built into the component; the divider’s ratio exists only while current flows through both sections.

Jason Yang's Blog

Physics Error Book

Force and Motion

Radioactivity

Electricity and Magnetism