Second practice final exam, fully corrected (PHYS 101)

Formula sheet, General equations: $|\vec a+\vec b|^{2} = a^{2}+b^{2}+2ab\cos\theta$, where $\theta$ is the interior angle between the two vectors drawn from the same point, exactly the $120^{\circ}$ of the figure.

$|\vec F|^{2} = 300^{2}+300^{2}+2(300)(300)\cos 120^{\circ} = 90000+90000-90000 = 90000$, so $|\vec F| = 300$ N.

Check by components along the dashed bisector: each rope gives $300\cos 60^{\circ} = 150$ N along it, the sideways parts cancel, and $150+150 = 300$ N.

Pitfall: $520$ N uses $\cos 60^{\circ}$, the angle of the triangle drawn head to tail, which is the supplement of the angle between the vectors. The angle of the sheet is the one between the two vectors placed tail to tail.

$600$ N adds the magnitudes, which is right only when the two ropes pull the same way.

$424$ N is Pythagoras, $300\sqrt 2$, right only at $90^{\circ}$.

$0$ N would need the two ropes to pull in opposite directions, at $180^{\circ}$.

$|\vec F|^{2} = 300^{2}+300^{2}+2(300)(300)\cos 60^{\circ} = 180000+90000 = 270000$, so $|\vec F| = 519.6$ N, about $520$ N.

The $520$ N of the wrong answer in a) belongs here: closing the angle between two pulls makes their sum larger.

Formula sheet, Work and energy: with no air resistance only gravity does work, so $KE + PE$ is the same on the bridge and at the water: $\frac{1}{2}mv^{2} = \frac{1}{2}mv_{0}^{2} + mgh$.

$v = \sqrt{v_{0}^{2}+2gh}$: neither the direction of the throw nor the mass appears.

$v = \sqrt{9.0^{2}+2(9.81)(12.0)} = \sqrt{81.0+235.4} = 17.8$ m/s for both stones.

Stone 1 does stay longer in the air, but it first climbs and loses speed: when it falls back through the level of the bridge it has $9.0$ m/s again, pointing $30^{\circ}$ below the horizontal, exactly like stone 2.

Stone 2 heads down from the start, but with the same speed and the same $12.0$ m to fall, so it gains the same energy.

The masses cancel in the energy balance, so no information is missing.

Vertical motion, upward positive, with the constant acceleration line of the formula sheet and the quadratic formula: $-12.0 = v_{0y}t - 4.905t^{2}$.

Stone 2, $v_{0y} = -4.5$ m/s: $4.905t^{2}+4.5t-12.0 = 0$ gives $t = \frac{-4.5+\sqrt{4.5^{2}+4(4.905)(12.0)}}{2(4.905)} = \frac{-4.5+15.99}{9.81} = 1.17$ s.

Stone 1, $v_{0y} = +4.5$ m/s: $4.905t^{2}-4.5t-12.0 = 0$ gives $t = \frac{4.5+15.99}{9.81} = 2.09$ s.

The difference, $0.92$ s, is exactly the time stone 1 spends going up and coming back to the level of the bridge, $\frac{2v_{0y}}{g} = \frac{2(4.5)}{9.81} = 0.92$ s.

Pitfall: the same speed at the water does not mean the same time in the air.

The two forces are the two members of one action-reaction pair: Newton's third law makes them equal in magnitude and opposite in direction at every instant, whatever the masses and whatever the motion.

The pair speeds up because of the forces acting ON it from outside, chiefly the road on the tyres, not because one member of a pair beats the other.

The car pushes harder: this compares two forces that act on different bodies. What decides the acceleration of the pickup is the net force on the pickup alone, the push of the car minus the resistance on the pickup.

The pickup pushes harder: this confuses a force with its effect. The same force gives the heavier pickup the smaller acceleration.

Equal only at constant speed: that is the first law dressed as the third. The third law holds while accelerating too.

The engine turns the wheels; the tread pushes BACKWARD on the road, and the road pushes FORWARD on the tread, by the third law again. That forward push is the only horizontal external force that can speed up the car and the pickup.

The tread does not slip on the road, so the friction is STATIC: whatever is needed, up to $\mu_{s}F_{N}$ (formula sheet, Dynamics: $F_{fr,s} \le \mu_{s}F_{N}$).

Pitfall: friction opposes the SLIPPING of the two surfaces, not the motion of the car. That is why it can point forward.

None, the engine pushes: the engine is inside the car, and internal forces cannot accelerate a body. On glare ice the wheels spin and the car stays put.

Kinetic friction would mean the tyres are spinning on the road, which is exactly what a driver avoids.

Formula sheet, Circular motion and gravity: $\frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM}$. Both moons orbit the same planet, so the right side is the same number for both, and the unknown mass of the planet cancels in a ratio.

$\left(\frac{T_{2}}{T_{1}}\right)^{2} = \left(\frac{r_{2}}{r_{1}}\right)^{3} = 4.0^{3} = 64$, so $\frac{T_{2}}{T_{1}} = 8.0$ and $T_{2} = 16$ days.

$8.0$ days takes $T \propto r$; $4.0$ days takes $T \propto \sqrt{r}$; $32$ days takes $T \propto r^{2}$; $128$ days takes $T \propto r^{3}$ and forgets the square root of the left side. Only $T^{2} \propto r^{3}$ is on the sheet.

On a circle, speed is circumference over period, $v = \frac{2\pi r}{T}$.

$\frac{v_{1}}{v_{2}} = \frac{r_{1}}{r_{2}}\cdot\frac{T_{2}}{T_{1}} = \frac{1}{4.0}(8.0) = 2.0$: the inner moon is twice as fast.

Same result from gravity alone: $\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$ gives $v = \sqrt{\frac{GM}{r}}$, and $\sqrt{4.0} = 2.0$.

Pitfall: the longer path of the outer moon does not make it faster. It is slower AND has farther to go, which is why its period is eight times longer.

Formula sheet, Work and energy: $W = \vec F\cdot\vec d = Fd\cos\theta$. The rope pulls UP while the piano moves DOWN, so $\theta = 180^{\circ}$ and the work is negative.

Constant speed means zero acceleration, so the tension equals the weight: $T = mg = 90(9.81) = 883$ N.

$W_{rope} = Td\cos 180^{\circ} = -883(6.0) = -5.30\times 10^{3}$ J, that is $-mgh = -5.30$ kJ.

Positive work would mean the rope helps the motion; here it holds the piano back, which is the whole point of the rope.

Zero work confuses the net work with the work of one force: the speed is constant, so the TOTAL is zero, not each term.

Smaller in size than $mgh$ would need the piano to speed up on the way down, which it does not.

Formula sheet: $W_{net} = \Delta KE$, and the kinetic energy does not change, so $W_{net} = 0$.

Term by term: gravity does $+mgh = +5.30$ kJ, the rope $-5.30$ kJ, and the sum is zero. The $5.30$ kJ of potential energy the piano loses end up as heat in the hands of the movers and in the rope rubbing on the railing.

No time is needed: work is a force times a displacement.

Paddler and canoe form a system that is isolated horizontally and starts at rest. Formula sheet, Momentum and collisions: $p_{1,i}+p_{2,i} = p_{1,f}+p_{2,f}$, so the total momentum stays zero.

While she moves forward, the canoe must move backward with an equal and opposite momentum; when she stops, her momentum is zero, so the canoe's is zero too, and it stops.

It stays put: but she pushes the canoe backward with her feet, and nothing holds it.

It keeps drifting: that would leave a nonzero total momentum, appearing from zero with no external force.

It moves forward: the two momenta would then point the same way, and their sum could not be zero.

A total momentum of zero from rest means the centre of mass does not move: formula sheet, $x_{COM} = \frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$ keeps its value.

Let the canoe move back a distance $s$. Relative to the water she moves forward $3.0 - s$, and the shifts of the two masses must balance: $70(3.0 - s) = 50s$.

$s = \frac{70(3.0)}{70+50} = 1.75$ m, and she moves $3.0 - 1.75 = 1.25$ m relative to the water.

Check: the heavier body moves less, and $70(1.25) = 50(1.75) = 87.5$ kg m.

Pitfall: the $3.0$ m is measured on the canoe, which is itself moving. Her displacement relative to the water is not $3.0$ m.

In the air the only external force is her weight, which acts at her centre of mass and has no torque about it. Formula sheet, Rotational motion: $\tau = \frac{\Delta L}{\Delta t}$, so with no torque $L = I\omega$ stays constant.

$I_{1}\omega_{1} = I_{2}\omega_{2}$ gives $\omega_{2} = \frac{12}{4.0}(1.2) = 3.6$ revolutions per second: three times less inertia, three times faster.

$1.2$ rev/s keeps $\omega$ instead of $L$; $0.40$ rev/s divides by the ratio instead of multiplying; $10.8$ rev/s multiplies by its square, as if $L$ were $I\omega^{2}$.

Formula sheet: $KE_{rot} = \frac{1}{2}I\omega^{2}$, which can be written $\frac{L^{2}}{2I}$. With $L$ fixed, dividing $I$ by three multiplies the energy by three.

Before: $\omega_{1} = 2\pi(1.2) = 7.540$ rad/s and $KE = \frac{1}{2}(12)(7.540)^{2} = 341$ J.

After: $\omega_{2} = 2\pi(3.6) = 22.62$ rad/s and $KE = \frac{1}{2}(4.0)(22.62)^{2} = 1023$ J, three times as much.

The extra $682$ J come from her muscles: pulling her arms and legs toward the axis while they are turning takes work, done inside the system. Angular momentum is conserved, kinetic energy is not.

Unchanged confuses the two quantities; times nine squares $\omega$ but forgets that $I$ shrank; divided by three is the inverse of the truth.

Formula sheet, Simple harmonic motion: $\omega_{spring} = \sqrt{k/m}$ and $\omega_{pendulum} = \sqrt{g/L}$, with $T = \frac{2\pi}{\omega}$.

No $g$ for the spring: on the Moon it hangs less stretched, but it oscillates about that new equilibrium with the same $k$ and the same $m$, so with the same period.

A $g$ for the pendulum: its restoring force is a part of the weight, so a weaker $g$ means a weaker restoring force and a slower swing.

Both longer treats the spring as if gravity drove it. Spring longer and pendulum unchanged swaps the two formulas. Pendulum shorter misreads $T \propto \frac{1}{\sqrt g}$: a smaller $g$ gives a LONGER period.

At fixed $L$, $T \propto \frac{1}{\sqrt g}$, so $T_{Moon} = 1.50\sqrt{\frac{9.81}{1.62}} = 1.50(2.461) = 3.69$ s.

Sense check: $g$ is about six times smaller on the Moon and $\sqrt 6 \approx 2.5$, so the period should be about two and a half times longer.

Formula sheet, Sound: $\beta = 10\log_{10}\frac{I}{I_{0}}$. A difference of two levels is the logarithm of a ratio of intensities: $\beta_{1}-\beta_{2} = 10\log_{10}\frac{I_{1}}{I_{2}}$.

$30 = 10\log_{10}\frac{I_{1}}{I_{2}}$ gives $\frac{I_{1}}{I_{2}} = 10^{3} = 1000$.

$30$ treats the decibel as a plain unit of intensity; $3$ stops at the exponent, $30/10$, and forgets the power of ten; $1.4$ divides the two levels as if they were proportional to the intensities, which a logarithmic scale never is.

The amplitude form of the same line of the sheet, $\beta = 20\log_{10}\frac{A}{A_{0}}$, says that the pressure amplitude is divided by $10^{30/20} = 31.6$: an intensity goes as the square of an amplitude.

$112 - 30 = 82$ dB, so $I = I_{0}10^{8.2} = (1.0\times 10^{-12})(1.585\times 10^{8}) = 1.58\times 10^{-4}$ W/m$^2$.

Check: at $112$ dB the intensity is $10^{11.2}I_{0} = 0.158$ W/m$^2$, and a thousandth of it is the same $1.58\times 10^{-4}$ W/m$^2$.

Formula sheet, Interference and diffraction: Rayleigh criterion $\theta = 1.22\frac{\lambda}{D}$. Two points are resolved when their angular separation is at least $\theta$, so a SMALLER $\theta$ is better: a shorter wavelength or a wider aperture.

Blue light has a shorter wavelength than red, so a smaller diffraction disc: this is the answer.

Infrared is LONGER than red: the discs grow and the resolution gets worse, even if haze bothers it less.

More pixels sample the blurred image more finely, but each point is still a disc of angular size $1.22\frac{\lambda}{D}$: the limit is set by the aperture, not by the detector.

Flying higher keeps the same $\theta$ but spreads it over a larger distance on the ground, $s = h\theta$: worse.

$\theta = 1.22\frac{500\times 10^{-9}}{0.50} = 1.22\times 10^{-6}$ rad.

Straight below, at a distance $h$, the angle is small, so $s = h\theta = (6.00\times 10^{5})(1.22\times 10^{-6}) = 0.73$ m.

Sense check: such a camera can tell two parked cars apart, not two people standing side by side.

Each pull is resolved on the canal axes, angles measured from $+x$: $+25^{\circ}$ for tug 1, $-40^{\circ}$ for tug 2.

$F_{1x} = 4.0\cos 25^{\circ} = 3.625$ kN and $F_{1y} = 4.0\sin 25^{\circ} = 1.690$ kN.

$F_{2x} = 3.0\cos 40^{\circ} = 2.298$ kN and $F_{2y} = -3.0\sin 40^{\circ} = -1.928$ kN.

$R_{x} = 5.923$ kN and $R_{y} = 1.690 - 1.928 = -0.238$ kN.

$R = \sqrt{5.923^{2}+0.238^{2}} = 5.93$ kN, at $\arctan\frac{0.238}{5.923} = 2.3^{\circ}$ to the RIGHT of the axis, since $R_{y} < 0$.

Pitfall: adding the magnitudes, $7.0$ kN, ignores the fact that the two tugs partly pull against each other across the canal.

Formula sheet, General equations: $|\vec a+\vec b|^{2} = a^{2}+b^{2}+2ab\cos\theta$, with $\theta$ the interior angle between the two pulls drawn from the bow: $25^{\circ}+40^{\circ} = 65^{\circ}$.

$R^{2} = 16.0+9.0+24.0\cos 65^{\circ} = 25.0+10.14 = 35.14$ kN$^{2}$, so $R = 5.93$ kN, as in a).

Using $115^{\circ}$, the angle of the head to tail triangle, would give $\sqrt{25.0-10.14} = 3.85$ kN, less than the larger pull: impossible for two pulls less than $90^{\circ}$ apart.

Formula sheet, Work and energy: $W = \vec F\cdot\vec d$, and the displacement $\vec d = (150\,;\,0)$ m lies along the axis.

$W_{1} = F_{1x}(150) = 3625(150) = 5.44\times 10^{5}$ J $= 544$ kJ, which is also $Fd\cos 25^{\circ}$.

$W_{2} = F_{2x}(150) = 2298(150) = 3.45\times 10^{5}$ J $= 345$ kJ.

Total: $888$ kJ, equal to $\vec R\cdot\vec d = 5923(150)$, as it must be: work adds like the forces.

The $y$ components, $1.690$ kN and $1.928$ kN, are perpendicular to the displacement: their dot product with $\vec d$ is zero, so they do no work. A tug pulling at a steep angle spends most of its effort holding the barge off the bank.

Constant velocity, so the net force is zero (Newton's first law) in each direction.

Along $x$: the drag is $5.92$ kN, pointing backward.

Along $y$: the water must push $+0.238$ kN, about $0.24$ kN, toward the left bank, to cancel $R_{y} = -0.238$ kN.

Net work: the drag does $-5923(150) = -888$ kJ, the sideways push none, so $W_{net} = 888 - 888 = 0$, which is the work energy theorem of the sheet, $W_{net} = \Delta KE = 0$ at constant speed. The $888$ kJ end up as heat and turbulence in the canal.

Formula sheet: power $= \vec F\cdot\vec v$, with $\vec v = (1.20\,;\,0)$ m/s.

$P_{1} = 3625(1.20) = 4.35\times 10^{3}$ W and $P_{2} = 2298(1.20) = 2.76\times 10^{3}$ W, total $7.11$ kW.

Check with c): the $150$ m take $\frac{150}{1.20} = 125$ s, and $\frac{888\times 10^{3}}{125} = 7.11\times 10^{3}$ W.

Axes: $x$ up the slope, $y$ perpendicular to it. Three forces: the weight $mg$, straight down; the normal force $F_{N}$, perpendicular to the snow; kinetic friction $F_{fr}$, DOWN the slope, because the board slides up it.

Along $y$, no acceleration: $F_{N} = mg\cos\theta$, so $F_{fr} = \mu_{k}mg\cos\theta$ (formula sheet, Dynamics).

Along $x$: $-mg\sin\theta-\mu_{k}mg\cos\theta = ma$, so the acceleration points down the slope with magnitude $a_{up} = g(\sin\theta+\mu_{k}\cos\theta)$.

$a_{up} = 9.81(0.2588+0.080\times 0.9659) = 9.81(0.3361) = 3.30$ m/s$^2$. The mass cancels.

Formula sheet, Kinematics: $v^{2} = v_{0}^{2}+2a(x-x_{0})$ with $v = 0$ at the highest point and $a = -a_{up}$:

Dimensions: $\frac{(\text{m/s})^{2}}{\text{m/s}^{2}} =$ m, a length, since sines, cosines and $\mu_{k}$ have no unit.

Limit $\mu_{k} \to 0$: $d = \frac{v_{0}^{2}}{2g\sin\theta}$, so the height gained is $d\sin\theta = \frac{v_{0}^{2}}{2g}$, all the kinetic energy turned into potential energy, as the energy line of the sheet predicts.

Variation: $\mu_{k}$ sits in the denominator, so more friction means a shorter climb, as it should.

$d = \frac{8.0^{2}}{2(3.297)} = 9.71$ m, against $12.6$ m without friction.

At the top she is momentarily at rest, and static friction could hold her only if $mg\sin\theta \le \mu_{s}mg\cos\theta$, that is $\tan\theta \le \mu_{s}$.

$\tan 15^{\circ} = 0.268 > 0.12$: the slope wins, she slides back.

On the way down friction points UP the slope, against the new sliding: $a_{down} = g(\sin\theta-\mu_{k}\cos\theta) = 9.81(0.2588-0.0773) = 1.78$ m/s$^2$.

Pitfall: keeping friction pointing down the slope after the turn around gives $3.30$ m/s$^2$ again, as if friction helped her slide.

Same line of the sheet, from rest over the distance $d$: $v_{back}^{2} = 2a_{down}d$. Replacing $d$ by its expression from b):

Dimensions: a speed times a pure number. Limit $\mu_{k} \to 0$: the square root becomes $1$ and she comes back at $v_{0}$, the frictionless round trip.

$v_{back} = 8.0\sqrt{\frac{0.1815}{0.3361}} = 8.0(0.7350) = 5.88$ m/s.

Energy check, per kilogram: she loses $\frac{1}{2}(8.0^{2}-5.88^{2}) = 14.7$ J/kg, and friction takes $\mu_{k}g\cos\theta(2d) = 0.758(19.41) = 14.7$ J/kg over the round trip.

$t_{up} = \frac{v_{0}}{a_{up}} = \frac{8.0}{3.297} = 2.43$ s, and from rest $d = \frac{1}{2}a_{down}t^{2}$ gives $t_{down} = \sqrt{\frac{2d}{a_{down}}} = \sqrt{\frac{2(9.71)}{1.781}} = 3.30$ s.

In letters, $\frac{t_{down}}{t_{up}} = \sqrt{\frac{a_{up}}{a_{down}}} = 1.36$: the same distance is covered with a smaller acceleration on the way down, because friction adds to gravity going up and subtracts from it coming down.

With $\mu_{k} \to 0$ the two accelerations are equal and so are the two times, the symmetry of a frictionless slope.

Only gravity does work, so $KE + PE$ is conserved (formula sheet, extended work energy principle with $W_{non-cons.} = 0$): $mgh = mg(2R)+\frac{1}{2}mv_{T}^{2}$, which gives $v_{T}^{2} = 2g(h-2R)$.

At T the centre of the loop is BELOW the car, so the net force points down: $mg+F_{N} = \frac{mv_{T}^{2}}{R}$ (formula sheet, $a = \frac{v^{2}}{r}$).

The track can push but never pull, so $F_{N} \ge 0$, that is $v_{T}^{2} \ge gR$.

$2g(h-2R) \ge gR$ gives $h \ge 2R+\frac{R}{2}$:

$h_{min} = \frac{5R}{2} = 2.5(0.20) = 0.50$ m

Dimensions: $R$ is a length, so $h_{min}$ is a length.

The mass multiplies every term of the energy balance and of Newton's law, so it cancels: a heavier car needs the same height. Variation: a loop twice as large needs a start twice as high.

Pitfall: $h = 2R$ is enough to REACH the height of the top only if the car could arrive there at rest, and at rest it cannot follow a curved path. See e).

At B the centre is ABOVE the car: $F_{N}-mg = \frac{mv_{B}^{2}}{R}$, with $v_{B}^{2} = 2gh$ from energy, so $F_{N} = mg\left(1+\frac{2h}{R}\right)$.

Limit $h \to 0$: $F_{N} = mg$, the car at rest at the bottom, as expected.

With $h = \frac{5R}{2}$: $F_{N} = mg(1+5) = 6mg = 6(0.050)(9.81) = 2.94$ N, six times the weight of the car.

$v_{T}^{2} = 2(9.81)(0.60-0.40) = 3.92$ m$^2$/s$^2$, and $F_{N} = m\left(\frac{v_{T}^{2}}{R}-g\right) = 0.050(19.62-9.81) = 0.49$ N.

In letters $F_{N} = mg\left(\frac{2h}{R}-5\right)$, and $\frac{2(0.60)}{0.20}-5 = 1$: at the top the track pushes down on the car with exactly its weight.

Check: at $h = h_{min}$ the same formula gives $F_{N} = 0$, the result of b).

Take a point of the upper half of the loop at an angle $\alpha$ above the horizontal through the centre: its height is $R(1+\sin\alpha)$.

Energy from $h = 2R$: $v^{2} = 2g\left(2R-R-R\sin\alpha\right) = 2gR(1-\sin\alpha)$.

Toward the centre act $F_{N}$ and the part $mg\sin\alpha$ of the weight: $F_{N}+mg\sin\alpha = \frac{mv^{2}}{R}$, so $F_{N} = mg(2-3\sin\alpha)$.

$F_{N}$ reaches zero when $\sin\alpha = \frac{2}{3}$, $\alpha = 41.8^{\circ}$, at a height $R\left(1+\frac{2}{3}\right) = \frac{5R}{3} = 0.333$ m, while the top is at $0.40$ m.

There the car leaves the track still moving, and falls across the inside of the loop as a projectile.