AP^® Physics 1 Flashcards

$\vec a = g = 9.8$ m/s$^2$ downward. The vertical velocity component $v_y$ is zero at the peak, but acceleration is unchanged: gravity does not turn off. This is a classic MC trap.

The graph has a local maximum or minimum: the slope $v = 0$, then changes sign. Equivalently, $v$ crosses zero with $a \neq 0$ on the $v$ vs $t$ graph.

Grows linearly. At time $t$ after A is dropped, A has fallen $\tfrac{1}{2} g t^2$; B (released at $t = 1$ s) has fallen $\tfrac{1}{2} g (t-1)^2$ for $t > 1$.

Gap $= \tfrac{1}{2} g [t^2 - (t-1)^2] = \tfrac{1}{2} g (2t - 1)$, which grows linearly in $t$.

• Distance: a scalar, the total path length, always $\geq 0$.
• Displacement: a vector, $\Delta \vec r = \vec r_f - \vec r_i$, which can be zero (round trip) or negative (in 1D).

For a 400 m runner finishing where they started, distance $= 400$ m but displacement $= 0$.

$f_s = 16.6$ N up the slope. Static friction balances the component of gravity along the slope: $f_s = m g \sin\theta = 4(9.8)(\sin 25°) = 4(9.8)(0.423) = 16.6$ N.

Note this is not $\mu_s F_N$: $f_s$ adjusts to whatever value is needed, up to its maximum.

Newton's third law: for every force from A on B, B exerts an equal and opposite force on A. The two forces act on different objects, so they never cancel each other on one object.

Common error: students think gravity on a book and the normal force from the table are a "third-law pair." They are not. Both act on the book; they happen to be equal and opposite, but for a different reason (Newton's second law: $\sum F = 0$).

It is not a separate force. It is the net force in the radial direction, supplied by real forces (tension, gravity, friction, normal). On an FBD you draw the real forces; centripetal "force" is just the label for their net inward sum, equal to $m v^2 / r$.

$g = 2.45$ m/s$^2$. Using $g(r) = G M_E / r^2$ at $r = 2 R_E$: $g = G M_E / (2 R_E)^2 = \tfrac{1}{4} g_{\text{surface}} = 9.8/4 = 2.45$ m/s$^2$. Gravity falls as $1/r^2$ from the center, not the surface.

$K$ increases by a factor of 9. Since $K \propto v^2$, tripling $v$ multiplies $K$ by $3^2 = 9$ (halving $v$ would quarter it). This non-linear relationship explains why doubling highway speed quadruples the kinetic energy your brakes must dissipate.

When the force is perpendicular to the displacement. Examples:

• Normal force on a sliding block (perpendicular to horizontal motion).
• Centripetal tension on a ball in horizontal circular motion (perpendicular to tangential velocity).
• Gravity on a satellite in circular orbit (perpendicular to orbital velocity).

A turning point is where $E = U(x)$, meaning $K = 0$ and the object momentarily stops. Beyond this $x$, $K$ would be negative (impossible), so the object reverses direction.

For a particle oscillating in a $U$-bowl, the turning points are the two values of $x$ where $U(x) = E$.

• Conservative (path-independent work, has an associated $U$): gravity, ideal spring force. Work to lift a ball 1 m up depends only on the 1 m change in height, not the route.
• Non-conservative (path-dependent, no $U$): friction, drag. Dragging a box across the floor and back to the start gives zero displacement but non-zero dissipated energy.

When a net external force acts on the system over a non-negligible time interval. Examples: a ball falling under gravity (gravity is external), or a block sliding with friction (friction is external).

For brief collisions, gravity's impulse $m g \Delta t$ is usually much smaller than the collision impulse, so momentum is conserved to good approximation.

They are equivalent: each doubles $\Delta p$. Since $\Delta p = F \Delta t$, a 200 N force for 0.1 s gives the same impulse as a 100 N force for 0.2 s. Peak force is double in the first case, so safety devices prefer the second.

$J = 2$ N s. Impulse equals the area under the curve: $\tfrac{1}{2} \text{base} \times \text{height} = \tfrac{1}{2}(0.08)(50) = 2$ N s.

Momentum conservation requires only that the net external force be zero during the interaction. Kinetic energy conservation additionally requires the collision force itself to be conservative, storing all its work in a recoverable form (like a perfect spring).

Real collisions deform materials and generate heat and sound, so $K$ leaks out: $p$ is preserved by Newton's third law, but $K$ is not.

$\tau = 5$ N m. $\tau = r F \sin 30° = (0.2)(50)(0.5) = 5$ N m, halved because only the perpendicular component of $\vec F$ contributes.

From smallest to largest (coefficient of $MR^2$):

• Solid sphere: $\tfrac{2}{5} = 0.40$.
• Solid cylinder: $\tfrac{1}{2} = 0.50$.
• Hollow sphere: $\tfrac{2}{3} \approx 0.67$.
• Hollow cylinder (hoop): $1$.

Mass concentrated near the axis means smaller $I$. The solid sphere wins a ramp race; the hoop loses.

To compute torques you need both the line of action and the application point of each force. Collapsing all forces to the COM (as in point-particle FBDs) loses the location information and makes torque calculation impossible. The COM convention is fine for $\sum F = m a_{\text{COM}}$ but not for $\sum \tau$.

The disc. For rolling without slipping, energy conservation gives $v^2 = 2gh/(1 + I/(MR^2))$.

• Disc: $I/(MR^2) = 0.5$, so $v^2 = 2gh/1.5 = (4/3) gh$.
• Hoop: $I/(MR^2) = 1$, so $v^2 = gh$.

The object with the smaller $I/(MR^2)$ always wins, whatever its mass or radius, because more of the available $PE$ converts to translational $K$.

At the contact point, the rolling surface has zero velocity relative to the ground. Friction acts against any tendency for relative motion there, but no relative motion is actually occurring, so the friction is static.

Static friction does no thermal-energy dissipation (no relative sliding), which is why energy is conserved in rolling-without-slipping problems.

Because $K = L^2 / (2I)$: with $L$ constant, a smaller $I$ means a larger $K$. The added $K$ comes from the work your muscles do pulling your arms inward. $L$ is conserved because there is no external torque; $K$ is not conserved.

Since $a = -\omega^2 x$:

• Maximum $|a|$ at $x = \pm A$ (the extremes).
• Zero $a$ at $x = 0$ (equilibrium).

This is the opposite of $|v|$, which is maximum at $x = 0$.

No. For ideal SHM, $T$ depends only on system parameters ($m$ and $k$ for a spring; $L$ and $g$ for a pendulum). Doubling the amplitude doubles $v_{\max}$ and $a_{\max}$ but leaves $T$ unchanged.

Real pendulums do show a small amplitude dependence at large angles, but small-angle SHM does not.

At the zero crossings of $x$, where the slope of $x(t)$ is largest in magnitude. At the peaks and troughs of $x$, the velocity is zero (the slope is flat). Same logic as the motion graphs in Unit 1.

About 89.5% submerged. For a floating object, $V_{\text{sub}}/V = \rho_{\text{ice}}/\rho_{\text{water}} = 917/1025 = 0.895$.

Each layer of fluid must support the weight of all the fluid above it. At depth $h$, a horizontal disc of fluid feels pressure from above equal to atmospheric plus the weight of the column above: $P_0 + \rho g h$. This is Newton's second law (no net force) applied to a fluid element.

Decreases. By Bernoulli's equation, $P + \tfrac{1}{2} \rho v^2 =$ constant (ignoring height changes). Faster flow ($v_2 > v_1$ by continuity) means lower $P$. This counterintuitive result explains the Venturi effect.