Corrected practice final exam in mechanics (203-SN1-RE, formerly 203-NYA)

Friction always opposes the MOTION, never the other forces. On the way up it therefore points down the slope and adds to the component of the weight: $a_{1}=g(\sin\theta+\mu\cos\theta)=6.00$ m/s². On the way down it points up the slope and subtracts: $a_{2}=g(\sin\theta-\mu\cos\theta)=2.00$ m/s². The motion is therefore brisker going up than coming down, which often surprises students.

Adding the two equations: $2g\sin\theta=8.00$, so $\sin\theta=\frac{8.00}{19.6}\approx 0.408$ and $\theta\approx 24.1°$. Subtracting them: $2\mu g\cos\theta=4.00$, so $\mu\cos\theta=\frac{4.00}{19.6}\approx 0.204$; with $\cos\theta\approx 0.913$ this gives $\mu\approx 0.224$. Two timing measurements, made without ever touching the slope, are enough to extract its geometry AND its surface condition.

Going up: $d=\frac{v_{0}^{2}}{2a_{1}}=\frac{144}{12.0}=12.0$ m, reached in $t_{1}=\frac{12.0}{6.00}=2.00$ s. Coming down the same distance: $t_{2}=\sqrt{\frac{2d}{a_{2}}}=\sqrt{\frac{24.0}{2.00}}\approx 3.46$ s. Total round trip: about $5.46$ s, the descent lasting markedly longer than the climb.

$v_{back}=a_{2}t_{2}=(2.00)(3.464)\approx 6.93$ m/s, against $12.0$ m/s at launch: the cart comes back at only 58 % of its initial speed. Writing $v_{0}^{2}=2a_{1}d$ and $v_{back}^{2}=2a_{2}d$, the distance cancels: $\frac{v_{back}^{2}}{v_{0}^{2}}=\frac{a_{2}}{a_{1}}=\frac{1}{3}$. The ratio therefore depends neither on the launch speed nor on the mass: it is a signature of the slope itself. It is also the fraction of the kinetic energy that is returned after the round trip.

Trajectory: $y=x\tan 30°-\frac{gx^{2}}{2v_{0}^{2}\cos^{2}30°}=0.5774\,x-0.01633\,x^{2}$. Hillside: a line of negative slope through the origin, $y=-x\tan 20°=-0.3640\,x$. The impact point is the intersection of the two, which requires counting heights NEGATIVELY below the launch point: this is the only delicate step in setting up the equations.

$0.5774x-0.01633x^{2}=-0.3640x$ gives, once the solution $x=0$ (the launch point) is discarded, $0.9413=0.01633x$, that is $x\approx 57.6$ m. Drop in height: $y=-0.3640(57.63)\approx -21.0$ m, so the projectile lands 21 metres lower. Distance along the hillside: $\frac{57.63}{\cos 20°}\approx 61.3$ m.

Time of flight: $t=\frac{x}{v_{0}\cos 30°}=\frac{57.63}{17.32}\approx 3.33$ s. At impact, $v_{x}=17.32$ m/s, unchanged, and $v_{y}=20.0\sin 30°-gt=10.0-32.6\approx -22.6$ m/s. Magnitude: $v=\sqrt{17.32^{2}+22.61^{2}}\approx 28.5$ m/s. Direction: $\tan\alpha=\frac{22.61}{17.32}$, about $52.5°$ BELOW the horizontal, much steeper than the hillside, which is only 20°: the projectile arrives almost head-on to the ground.

By energy: the projectile has dropped $21.0$ m, so $\frac{1}{2}mv^{2}=\frac{1}{2}mv_{0}^{2}+mg(21.0)$, hence $v=\sqrt{v_{0}^{2}+2g(20.97)}=\sqrt{400+411}\approx 28.5$ m/s, the same. If only the magnitude were asked for, the energy method would be by far the quicker: it needs neither the time of flight, nor the launch angle, nor even the trajectory, only the drop in height. It cannot, however, give the DIRECTION of the velocity, which is why the two approaches remain complementary.

The only horizontal force is the TENSION in the string, directed toward the nail: it, and it alone, plays the centripetal role. At the break, $T=\frac{mv^{2}}{r}$, so $v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(45.0)(0.800)}{0.500}}=\sqrt{72.0}\approx 8.49$ m/s. Angular speed: $\omega=\frac{v}{r}\approx 10.6$ rad/s, a frequency of $\frac{\omega}{2\pi}\approx 1.69$ revolutions per second, or a period of about $0.592$ s.

$a_{c}=\frac{v^{2}}{r}=\frac{72.0}{0.800}=90.0$ m/s², about $9.2\,g$. And yet the MAGNITUDE of the velocity does not change from one instant to the next: the motion is uniform circular motion. This considerable acceleration only turns the velocity vector, never lengthens it, because it stays perpendicular to it at every instant. Confusing acceleration with a change in the magnitude of the velocity is the most persistent error of the chapter.

The moment the string breaks, no horizontal force acts on the puck any more, since the table is smooth: the weight and the normal force cancel. By Newton's FIRST law, the puck therefore carries on in a straight line at constant speed, along the TANGENT to the circle at the break point, not along the radius. This is the decisive point: it goes straight ahead, not outward.

The path is the tangent from the break point, at $0.800$ m from the centre; it reaches the edge, at $1.30$ m from the centre, forming a right triangle in which the radius $0.800$ is one side and $1.30$ the hypotenuse. The distance travelled is therefore $\sqrt{1.30^{2}-0.800^{2}}=\sqrt{1.05}\approx 1.02$ m, covered in $t=\frac{1.025}{8.485}\approx 0.121$ s.

No force pushes the puck outward, either before or after the break. Before, the only force is the string pulling INWARD, and it is this pull that bends a path which, without it, would be straight. After, there is no force at all, and the puck simply does what any body left to itself does: it keeps going straight. The impression of being thrown outward comes from the point of view of someone turning with the puck: in that accelerated frame, a force has to be invented to account for the motion, and it is called centrifugal. It has no agent, no reaction, and it disappears as soon as the scene is described from the ground.

The TOTAL momentum is conserved, because no external horizontal force acts: the spring exerts on the two blocks internal forces that are equal and opposite. It is zero before, so zero after. MECHANICAL energy is also conserved here, since the surface is smooth and the spring ideal, but it changes FORM: the elastic energy becomes entirely kinetic. What is not conserved is the kinetic energy taken on its own before and after, since it was zero at the start.

Zero momentum: $m_{1}v_{1}=m_{2}v_{2}$, that is $2.00v_{1}=3.00v_{2}$ and so $v_{1}=1.50v_{2}$. Energy: $\frac{1}{2}(2.00)v_{1}^{2}+\frac{1}{2}(3.00)v_{2}^{2}=60.0$. Substituting: $2.25v_{2}^{2}+1.50v_{2}^{2}=3.75v_{2}^{2}=60.0$, so $v_{2}=4.00$ m/s and $v_{1}=6.00$ m/s. The light block leaves one and a half times faster, in the inverse ratio of the masses.

$E_{1}=\frac{1}{2}(2.00)(36.0)=36.0$ J and $E_{2}=\frac{1}{2}(3.00)(16.0)=24.0$ J, whose sum is indeed the $60.0$ J we started with. It is the LIGHT block that carries off the most energy, in the ratio $\frac{36}{24}=1.50$, exactly the INVERSE of the mass ratio. The reason lies in the form $E=\frac{p^{2}}{2m}$: the two blocks share the same momentum in magnitude, so the one with the smaller mass gets the most energy out of it. That is also why the recoil of a rifle is bearable while the bullet is lethal.

$2.00$ kg block: all its energy is dissipated by friction, $\mu mgd=E_{1}$, so $d=\frac{v_{1}^{2}}{2\mu g}=\frac{36.0}{4.90}\approx 7.35$ m. $3.00$ kg block: its kinetic energy becomes potential energy, $h=\frac{v_{2}^{2}}{2g}=\frac{16.0}{19.6}\approx 0.816$ m. Notice that neither the distance nor the height depends on the mass: both results can be read directly off the speeds.

On the smooth ramp, mechanical energy is conserved: $v=\sqrt{2gh}=\sqrt{2(9.8)(1.80)}=\sqrt{35.28}\approx 5.94$ m/s. We used conservation of ENERGY, which is legitimate as long as no friction acts. It stops being legitimate for the collision: a collision in which the bodies stick together is perfectly inelastic, it dissipates energy in deformation and sound, and applying it there would give a wrong answer. For the collision, only momentum is conserved.

$m_{1}v=(m_{1}+m_{2})v'$, so $v'=\frac{(4.00)(5.940)}{10.0}\approx 2.38$ m/s. The collision divided the speed by 2.5, exactly the ratio of the masses involved.

Before: $E=\frac{1}{2}(4.00)(35.28)=70.6$ J. After: $E'=\frac{1}{2}(10.0)(5.645)=28.2$ J. Loss: $42.3$ J, which is 60 % of the available energy, gone in an instant. In general, $\frac{E'}{E}=\frac{m_{1}}{m_{1}+m_{2}}$, so the fraction LOST is $\frac{m_{2}}{m_{1}+m_{2}}=\frac{6}{10}=0.600$: it depends neither on the speed nor on the starting height, only on the two masses. The heavier the target compared with the projectile, the more dissipative the collision, which is precisely what is asked of a car's crumple zone.

The remaining energy is dissipated by friction: $\frac{1}{2}(m_{1}+m_{2})v'^{2}=\mu(m_{1}+m_{2})gd$, where the total mass cancels. $d=\frac{v'^{2}}{2\mu g}=\frac{5.645}{3.92}\approx 1.44$ m. Note the chain of reasoning: energy, then momentum, then energy again. Choosing the right law at each phase is exactly what the exam tests, and what the chapter sets cannot test.

Before: $v=\sqrt{2gh_{0}}=\sqrt{2(9.8)(1.60)}=\sqrt{31.36}=5.60$ m/s. After: $v'=\sqrt{2gh_{1}}=\sqrt{2(9.8)(0.900)}=\sqrt{17.64}=4.20$ m/s. Ratio: $e=\frac{4.20}{5.60}=0.750$. This is called the coefficient of restitution, and it also equals $\sqrt{\frac{h_{1}}{h_{0}}}=\sqrt{0.5625}$, which saves computing the speeds.

Energy is proportional to height, so the ball keeps $\frac{h_{1}}{h_{0}}=e^{2}=0.5625$, that is 56 %, and loses 44 % at each contact. This energy has not vanished: it went into permanent deformation of the floor and the ball, into heating of the rubber and into sound, that thump which is very exactly mechanical energy carried off as an acoustic wave.

Each bounce multiplies the height by $e^{2}$: $h_{n}=h_{0}e^{2n}=1.60\,(0.5625)^{n}$. For $n=4$: $h_{4}=1.60(0.5625)^{4}\approx 0.160$ m, a tenth of the initial height after only four bounces.

The ball first falls through $h_{0}$, then each bounce adds a rise AND a fall of $h_{n}$: $D=h_{0}+2\sum_{n\geq 1}h_{0}e^{2n}=h_{0}+2h_{0}\frac{e^{2}}{1-e^{2}}$. The sum is a geometric series with ratio $e^{2}=0.5625$, less than 1, hence convergent. $D=1.60+2(1.60)\frac{0.5625}{0.4375}\approx 1.60+4.11\approx 5.71$ m. Infinitely many bounces, and yet a finite distance.

First fall: $t_{0}=\sqrt{\frac{2h_{0}}{g}}=\sqrt{0.3265}\approx 0.571$ s. The $n$th bounce lasts $2\sqrt{\frac{2h_{n}}{g}}=2t_{0}e^{n}$, since the height is multiplied by $e^{2n}$ and the time by its square root. Total: $t=t_{0}+2t_{0}\frac{e}{1-e}=0.5714+2(0.5714)\frac{0.750}{0.250}=0.5714+3.4286=4.00$ s exactly. The result deserves a pause: the ball makes INFINITELY many bounces, every one of them real, and yet everything is over after four seconds. This is the same paradox as Zeno's, resolved the same way: infinitely many ever smaller terms can have a finite sum. In practice, the last bounces come so close together that you hear a continuous buzz, that of a coin finishing its fall flat on a table.

The ANGULAR MOMENTUM of the merry-go-round plus child system is conserved about the axis, because the contact forces between the two are internal and the axle exerts no torque. Energy is not conserved: the child grabs on, hands and soles rub, and the catch is a perfectly inelastic collision. An object moving in a straight line does have an angular momentum about a point: $L=mvd$, where $d$ is the distance from the line to the axis. Here the child aims tangentially at the rim, so $d=R$ and $L_{child}=(30.0)(4.00)(2.00)=240$ kg·m²/s.

$L_{total}=I\omega_{0}+mvR=(250)(1.20)+240=300+240=540$ kg·m²/s. After the jump, the moment of inertia becomes $I'=250+mR^{2}=250+(30.0)(4.00)=370$ kg·m². So $\omega'=\frac{540}{370}\approx 1.46$ rad/s: the merry-go-round SPEEDS UP, because the child brought more angular momentum per unit of inertia than the merry-go-round had.

Before: $E=\frac{1}{2}(250)(1.20)^{2}+\frac{1}{2}(30.0)(4.00)^{2}=180+240=420$ J. After: $E'=\frac{1}{2}(370)(1.459)^{2}\approx 394$ J. Loss: about $26$ J, 6 % of the total, dissipated as heat in the rubbing of hands and feet, and in the deformation of the muscles absorbing the shock. It is small because the child arrived at a speed already close to that of the rim, $\omega_{0}R=2.40$ m/s against $4.00$ m/s: jumping on in the direction of rotation, and at the right speed, is precisely the way not to get hurt.

The braking torque produces $\alpha=\frac{\tau}{I'}=\frac{12.0}{370}\approx 0.0324$ rad/s². Time: $t=\frac{\omega'}{\alpha}=\frac{1.459}{0.03243}=45.0$ s. Angle turned through: $\theta=\frac{\omega'^{2}}{2\alpha}\approx 32.8$ rad, that is $\frac{32.84}{2\pi}\approx 5.23$ revolutions. The same angle can be obtained by energy, writing that the work of the torque, $\tau\theta$, dissipates all the rotational kinetic energy: $\theta=\frac{E'}{\tau}=\frac{394}{12.0}\approx 32.8$ rad, which confirms the result by an independent route.

Equilibrium requires the resultant force to be zero AND the sum of the torques to be zero. As long as the wind stays light, friction with the ground provides the first condition and the billboard does not slide; it is the second that gives way first, about the lower edge on the downwind side, as soon as the torque of the wind exceeds that of the weight. The billboard TIPS before it slides, which is the case whenever the object is tall and narrow.

Take torques about the downwind edge. The weight, applied at the centre, resists with a lever arm $\frac{b}{2}=0.400$ m: a torque of $(588)(0.400)=235$ N·m. The wind, applied at $\frac{H}{2}=1.20$ m, tends to tip it over. The limit is reached when $F(1.20)=235.2$, that is $F=\frac{mgb}{H}=\frac{(588)(0.800)}{2.40}=196$ N. That is little: a moderate gust over two square metres is plenty.

The centre of mass is initially at $\frac{H}{2}=1.20$ m; once the billboard is flat, it is at $\frac{b}{2}=0.400$ m. It therefore drops by $0.800$ m. Moment of inertia: $I=\frac{(60.0)(2.40^{2}+0.800^{2})}{3}=\frac{(60.0)(6.40)}{3}=128$ kg·m². Conservation of energy: $\frac{1}{2}I\omega^{2}=mg(0.800)$, so $\omega=\sqrt{\frac{2(588)(0.800)}{128}}=\sqrt{7.35}\approx 2.71$ rad/s. The upper edge is at $\sqrt{H^{2}+b^{2}}=\sqrt{6.40}\approx 2.53$ m from the pivot edge, so it arrives at $v=\omega r\approx 6.86$ m/s, nearly 25 km/h. A falling billboard is not a harmless object.

$F=\frac{(588)(1.20)}{2.40}=294$ N. Widening the base by 50 % raises the threshold by 50 % too: the relation $F=\frac{mgb}{H}$ is strictly proportional to $b$. The measure is therefore effective but costly in footprint, and above all linear: to double the wind resistance, the base must be doubled. Making the billboard lighter would not help, quite the opposite, since $F$ is proportional to $m$: it is the weight that holds it. The two real solutions are therefore to add ballast or to anchor it, which is exactly what we see on construction-site signs, systematically loaded with concrete blocks at their base.

The bucket does not fall freely because the rope holds it back: it exerts an upward tension that is not zero, since a torque is needed to set the winch turning. A fall at $g$ would require zero tension, hence a winch with no inertia. The equations: for the bucket, $mg-T=ma$; for the winch, $TR=I\alpha$. Since the rope does not slip, it imposes $a=\alpha R$, which links the two.

Substituting $\alpha=\frac{a}{R}$ into the second equation: $T=\frac{Ia}{R^{2}}$. Putting this into the first: $mg=a\left(m+\frac{I}{R^{2}}\right)$, with $I=\frac{1}{2}(40.0)(0.300)^{2}=1.80$ kg·m² and $\frac{I}{R^{2}}=20.0$ kg. So $a=\frac{(15.0)(9.8)}{15.0+20.0}=\frac{147}{35.0}=4.20$ m/s², only 43 % of $g$. Tension: $T=m(g-a)=(15.0)(5.60)=84.0$ N, well below the weight of $147$ N, as it must be since the bucket accelerates downward. The term $\frac{I}{R^{2}}$ acts exactly like an extra 20 kg of mass that has to be set in motion.

$v=\sqrt{2ad}=\sqrt{2(4.20)(6.00)}=\sqrt{50.4}\approx 7.10$ m/s. Angular speed: $\omega=\frac{v}{R}\approx 23.7$ rad/s. Number of turns: the unwound rope is $6.00$ m long and one turn unwinds $2\pi R=1.885$ m, hence $\frac{6.00}{1.885}\approx 3.18$ turns.

The potential energy lost is $mgh=(15.0)(9.8)(6.00)=882$ J. It is shared between the bucket and the winch: $\frac{1}{2}mv^{2}=\frac{1}{2}(15.0)(50.4)=378$ J and $\frac{1}{2}I\omega^{2}=\frac{1}{2}(1.80)(560.0)\approx 504$ J. Sum: $882$ J, the books balance exactly. Notice that the winch carries off 57 % of the energy, more than the bucket itself: this is the energy translation of the fact that its equivalent inertia, 20 kg, exceeds the mass of the bucket.

At constant speed, the acceleration is zero: the bucket is in equilibrium, so $T=mg=147$ N, and so is the winch, so the braking torque must exactly cancel that of the rope: $\tau=TR=(147)(0.300)=44.1$ N·m. It is this torque, not a force, that the brake of a real winch must supply, and it is independent of the descent speed chosen.

Count the direction of the probe's initial motion as positive. The planet moves at $-13.0$ km/s. In the planet's frame, the speed of the probe is $10.0-(-13.0)=23.0$ km/s: the probe arrives much faster than its solar speedometer says, and that is already half the explanation.

In a head-on elastic collision between a light body and a body of immensely greater mass, the light one leaves with the same speed in magnitude and the opposite direction, exactly like a ball bouncing off a wall. Here the mass of the planet exceeds that of the probe by twenty-four orders of magnitude: the approximation is excellent. The probe therefore leaves at $-23.0$ km/s in the planet's frame. Physically, this is of course not a bounce: it is gravity which, with no contact, bends the trajectory until its direction is reversed.

We return to the Sun's frame by adding the velocity of the planet: $-23.0+(-13.0)=-36.0$ km/s. The probe therefore leaves at $36.0$ km/s, in the direction opposite to the one it came from. Letting $v$ be the initial speed of the probe and $V$ that of the planet, we get the classic formula $v'=v+2V$: the probe gains TWICE the speed of the planet, whatever its own momentum. This manoeuvre, called a gravitational slingshot, allowed the Voyager probes to reach the escape speed of the solar system without carrying the corresponding fuel.

$\Delta E=\frac{1}{2}m(v'^{2}-v^{2})=\frac{1}{2}(800)\left[(3.60\times 10^{4})^{2}-(1.00\times 10^{4})^{2}\right]\approx 4.8\times 10^{11}$ J, the energy content of about ten tonnes of fuel. This energy comes from nowhere other than the ORBITAL energy of the planet: the probe-plus-planet system conserves its energy and its momentum, so the planet gives up exactly what the probe gains. Gravity merely acts as the go-between, creating nothing.

Momentum is conserved: $M\Delta V=-m\Delta v$, with $\Delta v=-36.0-10.0=-46.0$ km/s. So $|\Delta V|=\frac{(800)(4.60\times 10^{4})}{1.90\times 10^{27}}\approx 1.9\times 10^{-20}$ m/s. That is as good as nothing: it would take over a hundred times the age of the universe for the planet to drift even a single metre because of it. It is this colossal mass ratio that makes the manoeuvre possible: the probe draws on a reservoir so large that it never runs dry, and the impression of gaining energy for free comes only from forgetting to look at the other side of the books.

Along the arc, energy is conserved. The bob has risen by $h=L(1-\cos 60°)=1.25(0.500)=0.625$ m, so $v=\sqrt{2gh}=\sqrt{12.25}=3.50$ m/s.

A head-on elastic collision conserves both momentum and kinetic energy, which gives $v_{1}'=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v=\frac{-1.00}{5.00}(3.50)=-0.700$ m/s and $v_{2}'=\frac{2m_{1}}{m_{1}+m_{2}}v=\frac{4.00}{5.00}(3.50)=2.80$ m/s. The bob BOUNCES BACK, which is the signature of a collision with a heavier body. Energy check: $\frac{1}{2}(2.00)(0.490)+\frac{1}{2}(3.00)(7.84)=0.490+11.76=12.25$ J, exactly the energy before the collision, $\frac{1}{2}(2.00)(12.25)$. Momentum check: $(2.00)(-0.700)+(3.00)(2.80)=-1.40+8.40=7.00$, equal to $(2.00)(3.50)$.