PHYS 101 practice midterm with full solutions (McGill)

The sign of $\cos\theta$ decides, and nothing else (formula sheet, General equations, vectors with interior angle $\theta$):

Here $\vec A\cdot\vec B = (5.0)(3.0)\cos 120^{\circ} = -7.5$: the difference is the longer one.

The first option is the reflex carried over from numbers. A sum of vectors is longer than the difference only when the angle is below $90^{\circ}$; here $\vec A + \vec B$ is even shorter than $\vec A$ alone.

The third option forgets that $-\vec B$ points the other way: the angle between $\vec A$ and $-\vec B$ is $60^{\circ}$, not $120^{\circ}$, so the two constructions are not alike.

The fourth option: the axes play no part. A length depends only on $A$, $B$ and the angle between them, and the figure gives all three.

Sum, with $\theta = 120^{\circ}$:

$|\vec A + \vec B|^{2} = 25 + 9 + 2(5.0)(3.0)\cos 120^{\circ} = 34 - 15 = 19$, so $|\vec A + \vec B| = 4.36$

Difference: $\vec A - \vec B = \vec A + (-\vec B)$, and the angle between $\vec A$ and $-\vec B$ is $180^{\circ} - 120^{\circ} = 60^{\circ}$:

$|\vec A - \vec B|^{2} = 25 + 9 + 2(5.0)(3.0)\cos 60^{\circ} = 49$, so $|\vec A - \vec B| = 7.00$

Check: both lie between $A - B = 2.0$ and $A + B = 8.0$, as any combination of these two magnitudes must.

Pitfall: in the sheet's formula, $\theta$ is the angle between the vectors placed TAIL TO TAIL. Feeding it the $60^{\circ}$ angle of the head-to-tail triangle swaps the two answers.

The distance from the start grows as long as the velocity is positive, whatever its size.

Between Q and R the scooter slows down but still moves forward: every second still adds distance. After R the velocity is negative and the scooter comes back.

The farthest point is therefore R, where $v$ crosses zero, not where $v$ is largest.

Q is the reflex of reading a velocity graph as if it were a position graph: the top of $v(t)$ is the fastest instant, not the farthest one.

P only marks the end of the speeding up; the scooter keeps moving forward for eight more seconds.

S would be right only if the velocity stayed positive to the end; between R and S the scooter rolls back by $3$ m.

The displacement is the area between the graph and the time axis, counted with its sign, cut into triangles and a rectangle:

$\Delta x_{0\to R} = \tfrac{1}{2}(4)(6) + (4)(6) + \tfrac{1}{2}(4)(6) = 12 + 24 + 12 = 48$ m

Check on the last piece with the formula sheet (Kinematics, constant acceleration): from Q to R, $a = \frac{0 - 6}{4} = -1.5$ m/s$^2$ and $v^{2} = v_{0}^{2} + 2a\,\Delta x$ gives $\Delta x = \frac{0 - 36}{2(-1.5)} = 12$ m, the area of the last triangle.

At S the scooter is back to $48 - \tfrac{1}{2}(2)(3) = 45$ m from its start.

Take down as positive and let $t$ be the time since the first drop left. For $t \ge 0.20$ s:

$v_{1} = gt$ and $v_{2} = g(t - 0.20)$, so $v_{1} - v_{2} = g(0.20)$

The relative velocity is constant and positive (formula sheet, Kinematics, reference frames): seen from the second drop, the first one moves away at a steady speed. The gap grows by the same amount every second, so it increases steadily.

Pitfall: the same acceleration keeps the DIFFERENCE of the velocities fixed, not the gap. That difference is not zero, because the first drop had a $0.20$ s head start. This is the second option's error.

The third option has it backwards: the second drop is always the slower one and can never catch up.

The fourth option would need the first drop to stop accelerating, as it does at terminal speed with air drag; this course neglects the air.

$v_{1/2} = v_{1} - v_{2} = (9.81)(0.20) = 1.96$ m/s, directed down.

Check with positions: at $t = 0.50$ s the drops have fallen $\tfrac{1}{2}(9.81)(0.50)^{2} = 1.226$ m and $\tfrac{1}{2}(9.81)(0.30)^{2} = 0.441$ m, a gap of $0.785$ m.

At $t = 0.60$ s the gap is $1.766 - 0.785 = 0.981$ m: it grew by $0.196$ m in $0.10$ s, that is $1.96$ m/s.

The time of flight belongs to the vertical column alone. Rising to $H$ and falling back from it takes, for each ball,

$t = 2\sqrt{\frac{2H}{g}} = 2\sqrt{\frac{2(15)}{9.81}} = 3.50$ s

The three balls share $H$, hence the same vertical launch speed $v_{y0} = \sqrt{2gH} = 17.2$ m/s and the same clock: all three spend the same time in the air.

The first option mixes the two columns: a longer range comes from a larger horizontal speed, not from more time.

The second option is right only when the heights differ (at the same speed, a steeper kick goes higher); here the heights are equal.

The fourth option: the angles are not needed, the height carries all the vertical information.

Same time of flight, so the range sets the horizontal component: $v_{x} = R/t$ gives $8.58$, $12.9$ and $17.2$ m/s.

The launch speed is $v_{0} = \sqrt{v_{x}^{2} + v_{y0}^{2}}$ with the same $v_{y0}$ for all three, so it is largest for ball 3: $24.3$ m/s, against $19.2$ m/s for ball 1 and $21.4$ m/s for ball 2.

Ball 3 leaves at exactly $45^{\circ}$, since $v_{x} = v_{y0}$. Check with the range equation of the formula sheet: $R = \frac{v_{0}^{2}\sin 2\theta}{g} = \frac{24.26^{2}}{9.81} = 60.0$ m.

Formula sheet, Kinematics, reference frames: $\vec v_{\text{plane rel. ground}} = \vec v_{\text{plane rel. air}} + \vec v_{\text{air rel. ground}}$.

Out, against the wind: $200 - 40 = 160$ km/h. Back, with the wind: $200 + 40 = 240$ km/h.

Both legs cover the same distance, so the plane spends MORE time at the slow speed than at the fast one: the time lost against the wind is not won back. The windy trip is longer.

Pitfall: the second option averages the two speeds, $\frac{160 + 240}{2} = 200$ km/h. Over equal DISTANCES the average speed is not the average of the two speeds.

The third option has no mechanism behind it: a wind along the route never shortens a round trip.

The fourth option: the order of the legs changes nothing, the two durations simply add.

$t = \frac{300}{160} + \frac{300}{240} = 1.875 + 1.250 = 3.125$ h, against $\frac{600}{200} = 3.000$ h on a calm day: $7.5$ minutes more.

In letters, with $L$ the distance, $v$ the airspeed and $w$ the wind: $t = \frac{L}{v - w} + \frac{L}{v + w} = \frac{2Lv}{v^{2} - w^{2}}$.

This is larger than $\frac{2L}{v}$ for every $w \ne 0$, equal to it when $w = 0$, and it grows without bound as $w \to v$: a wind as fast as the plane would keep it from ever reaching the town.

The knot is at rest, so the forces on it add up to zero (formula sheet, Dynamics, Newton's second law with $a = 0$).

Horizontally: $T_{1}\cos 30^{\circ} = T_{2}\cos 60^{\circ}$, so $\frac{T_{2}}{T_{1}} = \frac{\cos 30^{\circ}}{\cos 60^{\circ}} = 1.73$

The two horizontal pulls must cancel. The steeper cord turns fewer of its newtons sideways, so it needs more newtons in total: cord 2 has the larger tension.

The second option is the intuition that the flatter cord takes the strain; it is the steeper one that carries most of the weight.

The third option holds only for a symmetric hang; here the angles differ.

The fourth option: the mass scales both tensions, but their ratio depends on the angles alone.

Vertically: $T_{1}\sin 30^{\circ} + T_{2}\sin 60^{\circ} = mg = 58.9$ N

With $T_{2} = \sqrt{3}\,T_{1}$: $T_{1}\left(\tfrac{1}{2} + \tfrac{3}{2}\right) = 2T_{1} = 58.9$ N

$T_{1} = 29.4$ N and $T_{2} = 51.0$ N

Check: $T_{1} + T_{2} = 80.4$ N, MORE than the weight. Tensions are not shares of the weight: only their vertical components add up to $mg$.

Vertically the eraser is at rest: $F_{fr,s} = mg = 0.050(9.81) = 0.491$ N, upward.

Horizontally the push is balanced by the normal force of the board: $F_{N} = 3.0$ N.

Doubling the push doubles $F_{N}$, hence the MAXIMUM friction $\mu_{s}F_{N}$, from $1.2$ N to $2.4$ N. The friction actually exerted stays $0.491$ N: it only ever has to hold the weight.

Pitfall: the options that double the friction or set it to $2.4$ N read the formula sheet's $F_{fr,s} \le \mu_{s}F_{N}$ as an equality. Static friction is as large as it needs to be, never more.

The option that halves it has no source: nothing in the vertical balance changed.

The eraser starts to slide when the friction needed reaches its maximum, $\mu_{s}F_{N} = mg$:

$F_{min} = \frac{mg}{\mu_{s}} = \frac{0.491}{0.40} = 1.23$ N

Sense check: a slipperier board, with a smaller $\mu_{s}$, needs a harder push, as it should.

Constant speed on a circle is not constant velocity: the direction turns, so there is an acceleration $a = \frac{v^{2}}{r}$ pointing to the centre (formula sheet, Circular motion and gravity). III is true, I is false.

$v = \frac{2\pi r}{T} = \frac{2\pi(12)}{30} = 2.51$ m/s and $a = \frac{v^{2}}{r} = \frac{2.51^{2}}{12} = 0.526$ m/s$^2$

At the bottom the centre is above her: $F_{N} - mg = ma$, so $F_{N} = m(g + a)$. At the top the centre is below her: $mg - F_{N} = ma$, so $F_{N} = m(g - a)$. II is true, IV is false.

The option with III alone misses II. The option with I and IV picks the two false ones: I takes the speed for the velocity, IV adds an outward push to the top equation where none exists. The options with II, III and IV or with all four keep at least one of those errors.

$F_{N,\text{top}} = m(g - a) = 60(9.81 - 0.526) = 557$ N, against $m(g + a) = 620$ N at the bottom and a weight of $589$ N.

Check: bottom minus top is $2ma = 63$ N, small because $a$ is only about $5$ per cent of $g$: the rider barely feels the difference.

Up positive, origin at the floor, $t = 0$ at the release (formula sheet, Kinematics, $x = x_{0} + v_{0}t + \frac{1}{2}at^{2}$ with $a = -g$):

$y_{\text{hoop}} = H - \tfrac{1}{2}gt^{2}$ and $y_{\text{ball}} = ut - \tfrac{1}{2}gt^{2}$

They meet when $y_{\text{hoop}} = y_{\text{ball}}$. The two $\tfrac{1}{2}gt^{2}$ terms cancel and $H = ut_{m}$:

Dimensions: $\frac{[\text{m}]}{[\text{m/s}]} = [\text{s}]$, a time. Limit: when $u$ is very large, $t_{m} \to 0$, the ball reaches the platform almost at once, as it should. Sense of variation: a faster throw gives an earlier meeting.

$t_{m} = \frac{24.0}{15.0} = 1.60$ s

Dimensions: $\frac{[\text{m/s}^2][\text{m}^2]}{[\text{m}^2/\text{s}^2]} = [\text{m}]$. Limit: when $u$ is very large, $y_{m} \to H$, the meeting happens at the platform.

$y_{m} = 24.0 - \frac{9.81(24.0)^{2}}{2(15.0)^{2}} = 24.0 - 12.56 = 11.4$ m

They meet in the air only if $y_{m} > 0$, that is $u^{2} > \frac{gH}{2}$:

$u_{min} = \sqrt{\frac{gH}{2}} = \sqrt{\frac{9.81(24.0)}{2}} = 10.8$ m/s

Check: at $u = u_{min}$, $t_{m} = \frac{H}{u_{min}} = \sqrt{\frac{2H}{g}}$, exactly the time the hoop takes to reach the floor. The two results agree.

Formula sheet, $v = v_{0} + at$: $v_{\text{ball}} = u - gt_{m} = u - \frac{gH}{u}$

$v_{\text{ball}} = 15.0 - \frac{9.81(24.0)}{15.0} = 15.0 - 15.70 = -0.70$ m/s

Negative: the ball has passed its highest point and is already falling, slowly. Its apex came at $t = \frac{u}{g} = 1.53$ s, before the meeting at $1.60$ s.

The ball would still be rising at the meeting only if $u > \frac{gH}{u}$, that is $u > \sqrt{gH} = 15.3$ m/s: $15.0$ m/s falls just short.

Pitfall: thrown upward does not mean moving upward at every later instant. The sign comes out of the computation, never out of the story.

Both bodies have the same acceleration, $-g$, so their relative acceleration is zero. Formula sheet, reference frames: $v_{\text{ball rel. hoop}} = v_{\text{ball rel. floor}} + v_{\text{floor rel. hoop}}$.

At $t = 0$ the hoop is at rest and the ball moves up at $u$; the relative velocity, $u$, never changes. Seen from the hoop, the ball comes straight at it at the constant speed $u$ and covers $H$ in $\frac{H}{u}$. That is why $g$ dropped out of a), and why the meeting time would be the same on the Moon.

At the meeting: $v_{\text{ball rel. hoop}} = -0.70 - (-15.70) = 15.0$ m/s, upward, exactly $u$.

Formula sheet, reference frames: $\vec v_{\text{buoy rel. water}} = \vec v_{\text{buoy rel. boat}} + \vec v_{\text{boat rel. water}}$, component by component:

$v_{x} = 8.0\cos 35^{\circ} + 5.0 = 6.55 + 5.0 = 11.55$ m/s

$v_{y} = 8.0\sin 35^{\circ} + 0 = 4.59$ m/s

$v = \sqrt{11.55^{2} + 4.59^{2}} = 12.4$ m/s, at $\arctan\frac{4.59}{11.55} = 21.7^{\circ}$ above the horizontal

Pitfall: adding the speeds, $8.0 + 5.0 = 13.0$ m/s, and keeping $35^{\circ}$. Only the horizontal component gains the boat's speed: seen from the lake, the throw is faster but FLATTER.

Vertical column, up positive, origin at the water (formula sheet, $x = x_{0} + v_{0}t + \frac{1}{2}at^{2}$):

The quadratic formula of the sheet, positive root:

$t = \frac{4.59 + \sqrt{4.59^{2} + 4(4.905)(2.5)}}{2(4.905)} = \frac{4.59 + 8.37}{9.81} = 1.32$ s

Greatest height: $v_{y} = 0$ at the top, and $v^{2} = v_{0}^{2} + 2a\,\Delta y$ gives $\Delta y = \frac{4.59^{2}}{2(9.81)} = 1.07$ m, that is $2.5 + 1.07 = 3.57$ m above the water.

The boat's speed appears in neither result: it only changes the horizontal column.

Horizontal column, constant velocity: $\Delta x = v_{x}t = 11.55(1.321) = 15.3$ m from the release point.

The boat, at $5.0$ m/s, has moved $5.0(1.321) = 6.61$ m in the same time.

Relative to the crew member: $15.265 - 6.606 = 8.66$ m ahead of her.

The crewmate's prediction, $8.0\cos 35^{\circ} \times 1.321 = 8.66$ m, is right in the frame of the boat. That frame moves at constant velocity, so in it the buoy is an ordinary projectile launched at $8.0$ m/s and $35^{\circ}$, with the same $g$ and the same time of flight.

In the frame of the water the distance is larger by the boat's own displacement, $6.61$ m. The prediction does not ignore the boat's speed by mistake: in the boat's frame that speed is zero. And sharing the time of flight does not make the two frames agree on the horizontal distance, since their horizontal speeds differ by $5.0$ m/s.

Axes on the pallet: $x$ up the slope, $y$ perpendicular to the roof. The rope pulls up the slope with $T$, and while the bucket is at rest, $T = mg$.

Perpendicular to the roof: $F_{N} = Mg\cos\theta$. Along the slope, with the friction $f$ counted positive up the slope: $mg + f - Mg\sin\theta = 0$, so $f = Mg\sin\theta - mg$.

Static friction holds as long as $|f| \le \mu_{s}F_{N}$ (formula sheet, Dynamics, $F_{fr,s} \le \mu_{s}F_{N}$).

Light bucket, pallet about to slide DOWN: four forces, the weight, the normal force, the tension up the slope and the friction up the slope at its maximum, $mg = Mg\sin\theta - \mu_{s}Mg\cos\theta$.

Heavy bucket, pallet about to slide UP: the same four forces, but the friction now points down the slope, $mg = Mg\sin\theta + \mu_{s}Mg\cos\theta$.

Dimensions: $\mu_{s}$ and the trigonometric ratios are pure numbers, so both ends are masses. Limit $\mu_{s} \to 0$: the range shrinks to the single value $m = M\sin\theta$, the frictionless balance. Limit $\theta = 90^{\circ}$: a vertical wall, $\cos\theta = 0$, and $m = M$, two equal masses over a pulley; friction has nothing to press on.

$\sin 32.0^{\circ} = 0.5299$ and $\cos 32.0^{\circ} = 0.8480$:

$m_{min} = 40.0(0.5299 - 0.45 \times 0.8480) = 40.0(0.1483) = 5.93$ kg

$m_{max} = 40.0(0.5299 + 0.3816) = 40.0(0.9115) = 36.5$ kg

Pitfall: giving a single answer. Static friction gives a RANGE, because it can point either way and take any value up to its maximum.

$15.0$ kg lies inside the range: the pallet stays put and the friction is static, with whatever value equilibrium requires.

$f = Mg\sin\theta - mg = 40.0(9.81)(0.5299) - 15.0(9.81) = 207.9 - 147.2 = 60.8$ N

Positive, so up the slope: the bucket alone ($147$ N) cannot hold the pull of the weight down the slope ($208$ N), and friction supplies the rest.

Pitfall: writing $\mu_{s}F_{N} = 0.45(332.8) = 150$ N. That is the most friction can do, not what it does here.

Friction is not zero just because nothing moves: it is what keeps the pallet from moving. And a friction force always lies along the surface, never perpendicular to it.

$45.0$ kg is above $m_{max}$: the pallet moves up the slope and kinetic friction, $\mu_{k}F_{N}$, points down it. Newton's second law for each body along its own motion, with the same $a$ and the same $T$:

Bucket: $mg - T = ma$. Pallet: $T - Mg\sin\theta - \mu_{k}Mg\cos\theta = Ma$.

Adding the two equations eliminates $T$:

$a = \frac{45.0 - 21.20 - 10.18}{85.0}(9.81) = 1.57$ m/s$^2$

$T = m(g - a) = 45.0(9.81 - 1.57) = 371$ N, less than the bucket's weight of $441$ N, as it must be for the bucket to speed up downward.

Checks: a ratio of masses times $g$ is an acceleration. A larger $\mu_{k}$ makes the numerator smaller, hence a smaller acceleration, as a rougher roof should. With $\mu_{k} = 0$ and $\theta = 0$, it becomes $\frac{m}{m + M}g$, the familiar result for a block on a smooth table pulled by a hanging mass.

Formula sheet, Force of gravity: $F = \frac{Gm_{1}m_{2}}{r^{2}}$. At P:

$G$ and $m_{c}$ cancel. Both distances are positive, so taking square roots: $\frac{D - x}{x} = \sqrt{\frac{m}{M}}$, and

Dimensions: $\sqrt{m/M}$ is a pure number, so $x$ is a length. Limits: as $m \to 0$, $x \to D$, the point slides into a Moon that no longer pulls; with $m = M$, $x = \frac{D}{2}$, the midpoint, by symmetry.

$\sqrt{\frac{7.35\times10^{22}}{5.97\times10^{24}}} = 0.1110$, so $x = \frac{3.84\times10^{8}}{1.1110} = 3.46\times10^{8}$ m, that is $90.0$ per cent of the way.

Pitfall: setting $\frac{x}{D - x} = \frac{M}{m}$ forgets the square of the inverse-square law and puts P at $98.8$ per cent of the way.

Halfway, both centres are $\frac{D}{2} = 1.92\times10^{8}$ m away, and the two pulls point in opposite directions:

$F_{net} = \frac{Gm_{c}(M - m)}{(D/2)^{2}} = \frac{6.67\times10^{-11}(2.60\times10^{4})(5.90\times10^{24})}{(1.92\times10^{8})^{2}} = 277$ N, towards the Earth

Halfway is NOT the balance point: the Earth is $81$ times more massive and wins easily there. The pulls cancel only at P, much closer to the Moon.

Formula sheet, Circular motion: $v = \frac{2\pi D}{T}$ and $a = \frac{v^{2}}{D} = \frac{4\pi^{2}D}{T^{2}}$, with $T = 27.3(86400) = 2.359\times10^{6}$ s:

$a = \frac{4\pi^{2}(3.84\times10^{8})}{(2.359\times10^{6})^{2}} = 2.72\times10^{-3}$ m/s$^2$

Inverse-square prediction: $g\left(\frac{R_{E}}{D}\right)^{2} = 9.81\left(\frac{6.37\times10^{6}}{3.84\times10^{8}}\right)^{2} = 2.70\times10^{-3}$ m/s$^2$

The two agree within $1$ per cent: the pull that makes an apple fall and the pull that bends the Moon's path are one force, weakened by the square of the distance. This is the test Newton made.

Check with $G$ directly: $\frac{GM}{D^{2}} = \frac{6.67\times10^{-11}(5.97\times10^{24})}{(3.84\times10^{8})^{2}} = 2.70\times10^{-3}$ m/s$^2$.

Formula sheet, Kepler's third law: $\frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM}$, the same for every body that orbits the Earth.

Doubling $r$ multiplies $r^{3}$ by $8$, so $T^{2}$ by $8$ and $T$ by $\sqrt{8} = 2\sqrt{2} = 2.83$:

$T' = 27.3(2.83) = 77.2$ days

Twice as long would make $T$ proportional to $r$, four times as long proportional to $r^{2}$. And the period does depend on the radius: what stays the same is the ratio $\frac{T^{2}}{r^{3}}$, not $T$.