Honors Physics Unit 5 Review: Motion in Two Dimensions

A vector has magnitude and direction, so you can't add vectors like plain numbers. Walking 4 m east and then 3 m north doesn't put you 7 m from where you started.

The head-to-tail method fixes this. Draw the first vector to scale, put the tail of the next vector at the head of the first, and keep chaining. The resultant runs from the tail of the first vector to the head of the last one. For the walk above, the resultant is 5 m long, pointing about 36.9° north of east.

The parallelogram method gives the same answer. Place two vectors tail to tail, complete the parallelogram, and draw the diagonal from the shared tail. It's handy when two forces pull on the same point.

To subtract, add the opposite. $\vec{A} - \vec{B}$ means flip $\vec{B}$ so it points the other way, then add it head to tail. Order doesn't matter for addition ($\vec{A} + \vec{B} = \vec{B} + \vec{A}$), but it does for subtraction.

Graphical answers depend on your scale (for example, 1 cm = 5 N) and how carefully you use the ruler and protractor. They're great for checking reasonableness, but the analytical method in 5.2 gives exact results.

The analytical method uses trig instead of a ruler. For a vector of magnitude $A$ at angle $\theta$ from the positive x-axis, the components are $A_x = A\cos\theta$ and $A_y = A\sin\theta$. You can also write the vector with unit vectors, $\vec{A} = A_x\hat{i} + A_y\hat{j}$, where $\hat{i}$ and $\hat{j}$ each have length 1.

To add or subtract, combine matching components. Then rebuild the resultant with $R = \sqrt{R_x^2 + R_y^2}$ and $\theta = \tan^{-1}(R_y/R_x)$.

Worked example. Add $\vec{A}$ = 10 m at 30° and $\vec{B}$ = 6 m at 120°.

1. $A_x = 10\cos 30° = 8.66$ m and $A_y = 10\sin 30° = 5.00$ m.
2. $B_x = 6\cos 120° = -3.00$ m and $B_y = 6\sin 120° = 5.20$ m.
3. $R_x = 8.66 + (-3.00) = 5.66$ m and $R_y = 5.00 + 5.20 = 10.20$ m.
4. $R = \sqrt{5.66^2 + 10.20^2} = \sqrt{136.1} \approx 11.7$ m.
5. $\theta = \tan^{-1}(10.20/5.66) \approx 61.0°$.

The calculator's $\tan^{-1}$ returns angles between $-90°$ and $90°$. If $R_x$ is negative, add 180° to get the direction in the second or third quadrant. If $R_x = 0$, the direction is 90° for positive $R_y$ or 270° for negative $R_y$. A zero vector has no direction.

A projectile moves under gravity alone after launch. Ignoring air resistance and using a constant downward gravitational acceleration, its path is a parabola if it has horizontal velocity. A vertically launched projectile follows a vertical line. You can analyze the motion along two axes linked by time.

Set the origin at the launch point and take upward as positive. Horizontally, $a_x = 0$, so $x = v_{0x}t$. Vertically, $a_y = -g$, so $y = v_{0y}t - \tfrac{1}{2}gt^2$ and $v_y = v_{0y} - gt$. For a launch at speed $v_0$ and angle $\theta$, start with $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$.

Worked example. A ball rolls off a 1.25 m high table at 3.0 m/s. Where does it land?

1. It leaves horizontally, so $v_{0y} = 0$ and $v_{0x} = 3.0$ m/s.
2. Use the vertical drop to find time: $1.25 = \tfrac{1}{2}(9.8)t^2$, so $t = \sqrt{2(1.25)/9.8} \approx 0.51$ s.
3. Use that time horizontally: $x = (3.0)(0.51) \approx 1.5$ m from the table's edge.

At the top of a path, $v_y = 0$ but the acceleration is still $g$ downward, and $v_x$ is unchanged. For launch and landing at the same height, use $R = \dfrac{v_0^2\sin 2\theta}{g}$. For an upward launch, the maximum height above the launch point is $h_{max} = \dfrac{v_0^2\sin^2\theta}{2g}$, even if the landing height differs. With the same launch speed and equal launch and landing heights, range peaks at 45°, and complementary angles like 30° and 60° give equal ranges.

On a ramp, tilt your axes so x runs along the slope and y runs perpendicular to it. Then only gravity needs splitting. The component down the slope is $mg\sin\theta$, and the component into the surface is $mg\cos\theta$. If no other force acts on the block, the normal force balances the perpendicular component of gravity, so $N = mg\cos\theta$, not $mg$. A quick check is that at $\theta = 0°$, the slope component is zero and $N = mg$, as on flat ground.

Friction depends on $N$. Kinetic friction is $f_k = \mu_k N$. Static friction adjusts to whatever is needed, up to $f_{s,max} = \mu_s N$. A block stays at rest if $\tan\theta \le \mu_s$.

Worked example. A box slides down a 30° ramp with $\mu_k = 0.20$. Find its acceleration.

1. Newton's second law along the slope gives $ma = mg\sin\theta - \mu_k mg\cos\theta$.
2. Mass cancels, so $a = g(\sin\theta - \mu_k\cos\theta)$.
3. $a = 9.8(0.500 - 0.20 \times 0.866) = 9.8(0.327) \approx 3.2$ m/s².

Mass doesn't change the acceleration, though a heavier box does feel larger normal and friction forces. Steeper angles increase acceleration twice over, because the slope component grows while the normal force, and with it friction, shrinks.

Simple harmonic motion (SHM) happens when the restoring force is proportional to displacement and points back toward equilibrium. For a spring, that's Hooke's law, $F = -kx$. The minus sign means the force always opposes the stretch or compression, which is why the object oscillates.

For an ideal horizontal spring with no friction, energy trades between kinetic and elastic potential energy. At the turning points, the mass stops and the energy is elastic potential, $\tfrac{1}{2}kx^2$. At equilibrium, speed is greatest and the energy is kinetic. The total stays $\tfrac{1}{2}kA^2$, where $A$ is the amplitude.

Period $T$ is the time for one full cycle, and frequency is $f = 1/T$. For a mass on an ideal spring with no damping, $T = 2\pi\sqrt{m/k}$. For a simple pendulum at small angles (under about 15°), $T \approx 2\pi\sqrt{L/g}$. A pendulum's restoring force is $mg\sin\theta$, the same slope-direction component of gravity you used on ramps.

Worked example. Find the approximate period of a 0.80 m simple pendulum on Earth for a small-angle swing.

1. $T = 2\pi\sqrt{0.80/9.8} = 2\pi\sqrt{0.0816}$.
2. $T = 2\pi(0.286) \approx 1.8$ s, so $f \approx 0.56$ Hz.

The ideal spring period does not depend on amplitude. The small-angle pendulum formula does not include amplitude, but a pendulum's actual period increases as its swing gets larger. A simple pendulum's period does not depend on its mass, but the spring period depends on the attached mass.