Newton's Second Law: Cartesian - Statics and Dynamics

1. During the phase while the package is slipping relative to the belt, what are the correct $x$- and $y$-components of the equation of motion for the package, and how long does this slipping phase last?
   Answer: $x$: $m\dot{v}_x = \mu_k mg$; $y$: $N - mg = 0$. Slipping lasts $t = v_b/(\mu_k g) \approx 1.02\text{ s}$, after which the package reaches belt speed; the friction force is positive in $+x$ during slip because the belt moves faster than the package. Whenever you see a conveyor belt problem, your first instinct should be to identify who is moving faster — the belt or the package — because that determines the direction of kinetic friction on the package. Here, the belt moves at $v_b = 3\text{ m/s}$ while the package starts at rest. The belt surface moves faster than the package, so kinetic friction acts on the package in the **$+x$ direction**, accelerating it forward. The equations of motion are straightforward: vertically, the package has no acceleration, giving $N - mg = 0$, so $N = mg$. Horizontally, the friction force is $f_k = \mu_k N = \mu_k mg$, giving $m\dot{v}_x = \mu_k mg$, which simplifies to $\dot{v}_x = \mu_k g \approx 2.943\text{ m/s}^2$. Slipping ends when the package reaches belt speed: $t = v_b/(\mu_k g) = 3/(0.3 \times 9.81) \approx 1.02\text{ s}$. This is exactly what A describes — making it correct. B is wrong on two counts: friction acts in $+x$ (not $-x$), and dividing by $\mu_k$ alone (without $g$) produces a nonsensical time with wrong units. C writes the $x$-equation correctly but incorrectly claims $N > mg$ in the $y$-direction — friction is horizontal and creates no vertical imbalance, so $N = mg$ exactly. D gets the equations right but invents a "factor of two" correction based on average relative velocity, which has no basis in Newton's second law; the acceleration is constant, so the time is simply $v_b/(\mu_k g)$. Study tip: Always draw a free-body diagram and ask "which surface is sliding relative to the other, and in which direction?" — that single question pins down the friction direction on every conveyor belt problem.
2. Which of the following is a correct statement about the particle's trajectory and the direction of motion as $t \to \infty$?
   Answer: The trajectory is a straight line in the direction of $\vec{v}_0$ for all $t$, and the speed decays exponentially; the direction of motion remains $\alpha$ from the $x$-axis throughout the motion because the drag force components are proportional to velocity components independently. When you see a drag force whose components depend only on their corresponding velocity components — $F_x = -b\dot{x}$ and $F_y = -b\dot{y}$ — the key insight is that the two equations of motion are completely decoupled. You can solve each independently: $m\ddot{x} = -b\dot{x}$ gives $\dot{x}(t) = v_0\cos\alpha\, e^{-bt/m}$, and identically $\dot{y}(t) = v_0\sin\alpha\, e^{-bt/m}$. Both components decay with the same time constant $\tau = m/b$. Since both shrink by the same exponential factor at every moment, the ratio $\dot{y}/\dot{x} = \tan\alpha$ remains constant for all $t$. The direction of motion never changes — the particle travels in a straight line at angle $\alpha$, with speed decaying as $v_0 e^{-bt/m}$. This confirms B as correct. Answer A is wrong because both components share the same decay rate — there is no differential decay, quadratic coupling, or any mechanism that would favor one axis over the other. C is wrong because the drag force here is purely speed-reducing (it acts antiparallel to velocity) and contains no cross-component coupling that could rotate the velocity vector — no spiraling occurs. D is wrong for the same decoupling reason: there is no interaction between the $x$- and $y$-equations, so curvature never arises regardless of $\alpha$. A useful study habit: whenever drag components are written as $F_i = -b\dot{q}_i$ separately, immediately recognize that the axes decouple — identical time constants mean direction is preserved.
3. A student differentiates the $x$-equation of motion to obtain a single ODE for $x(t)$. Which of the following is the correct fourth-order ODE for $x$ alone, and what is the general solution form?
   Answer: Differentiating $m\ddot{x} = ay$ twice and substituting $\ddot{y} = \frac{b}{m}x$ gives $\ddddot{x} = \frac{ab}{m^2}x$; with $a = b = 6$, $m = 2$ this becomes $\ddddot{x} = 9x$, whose characteristic roots are $r = \pm\sqrt{3},\,\pm i\sqrt{3}$, yielding $x(t) = C_1 e^{\sqrt{3}t} + C_2 e^{-\sqrt{3}t} + C_3\cos(\sqrt{3}t) + C_4\sin(\sqrt{3}t)$. When a force depends on both coordinates, you can often eliminate one variable by differentiating an equation of motion and substituting the other equation — turning a coupled system into a single higher-order ODE. That's exactly the skill being tested here. Start with Newton's second law for each coordinate: $m\ddot{x} = ay$ and $m\ddot{y} = bx$. Differentiate the $x$-equation twice to get $m\ddddot{x} = a\ddot{y}$. Now substitute $\ddot{y} = \frac{b}{m}x$ from the $y$-equation: $m\ddddot{x} = a \cdot \frac{b}{m}x$, which simplifies to $\ddddot{x} = \frac{ab}{m^2}x$. Plugging in $a = b = 6$, $m = 2$ gives $\ddddot{x} = 9x$. The characteristic equation is $r^4 = 9$, so $r^4 - 9 = 0$, yielding $r^2 = \pm 3$. The positive root gives $r = \pm\sqrt{3}$ (real), and the negative root gives $r = \pm i\sqrt{3}$ (imaginary). This confirms D is correct, with general solution $x(t) = C_1 e^{\sqrt{3}t} + C_2 e^{-\sqrt{3}t} + C_3\cos(\sqrt{3}t) + C_4\sin(\sqrt{3}t)$. A is wrong because $r^4 = 9$ has four roots — not just real ones. Equal values of $a$ and $b$ don't eliminate imaginary roots; you must solve $r^4 = 9$ fully. B is wrong because differentiating only once and then substituting $\ddot{y} = \frac{b}{m}x$ does allow full elimination — no numerical integration is needed. C is wrong because it incorrectly differentiates only once before substituting, producing $\ddddot{x} = \omega^2 \ddot{x}$ instead of the correct $\ddddot{x} = \omega^2 x$, which changes both the characteristic equation and the solution entirely. Study tip: When eliminating variables from coupled ODEs, always track which derivative you differentiate into and which equation you substitute — mixing levels is the most common error and produces a completely different characteristic equation.
4. Which expression correctly gives $\vec{F}_3$ that produces the required motion, and what constraint on $\theta$ ensures $\vec{F}_3$ has no $\hat{j}$ component?
   Answer: $\vec{F}_3 = F_0(1 - \cos\theta)\hat{i} + (ma_0 - F_0\sin\theta)\hat{j}$; the constraint $\sin\theta = ma_0/F_0$ eliminates the $\hat{j}$ component of $\vec{F}_3$, so the required $\hat{j}$ acceleration comes entirely from $\vec{F}_1$'s vertical component. When a problem gives you multiple forces and a specified net acceleration, your first move should always be Newton's second law: $\sum \vec{F} = m\vec{a}$. Here, the block must accelerate as $m a_0 \hat{j}$, meaning the net force must equal $m a_0 \hat{j}$ — zero in $\hat{i}$, and $m a_0$ in $\hat{j}$. Start by summing $\vec{F}_1$ and $\vec{F}_2$: their combined $\hat{i}$ component is $F_0\cos\theta - F_0 = F_0(\cos\theta - 1)$, and their $\hat{j}$ component is $F_0\sin\theta$. For the total net force to equal $ma_0\hat{j}$, you need $\vec{F}_3$ to supply whatever is missing. The $\hat{i}$ component of $\vec{F}_3$ must cancel $F_0(\cos\theta - 1)$, giving $F_0(1-\cos\theta)$. The $\hat{j}$ component must supply $ma_0 - F_0\sin\theta$. This confirms answer B: $\vec{F}_3 = F_0(1-\cos\theta)\hat{i} + (ma_0 - F_0\sin\theta)\hat{j}$. When $\sin\theta = ma_0/F_0$, the $\hat{j}$ component of $\vec{F}_3$ vanishes, and $\vec{F}_1$'s vertical component alone drives the acceleration. A incorrectly writes $\vec{F}_3$ as if it still contains a separate $ma_0\hat{j}$ term after combining, double-counting the vertical contribution. C gives the wrong $\hat{i}$ component — $F_0\cos\theta$ instead of $F_0(1-\cos\theta)$ — suggesting it only cancels $\vec{F}_2$ without accounting for $\vec{F}_1$'s horizontal part. D flips the sign of the $\hat{i}$ component and imposes an overly restrictive constraint, valid only for a specific case rather than the general solution. Your strategy: always resolve all known forces component by component first, then solve for the unknown force as the difference between the required net force and the sum of known forces.
5. A student wants to find the $x$-position when the particle returns to $y = 0$ (the range). Which approach correctly applies Newton's second law in Cartesian components to set up this problem?
   Answer: Split the trajectory into three phases: (1) ascent with $y < H$: $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; (2) above $H$: $m\ddot{x} = 0$, $m\ddot{y} = -mg$; (3) descent with $y < H$: $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; match velocity and position at each phase boundary to find total range. When a force depends on position (or any condition that changes during motion), you must identify every distinct phase where the equations of motion change — and apply the correct equations within each phase, matching velocity and position at the boundaries. Here, the wind force $F_w\hat{i}$ acts only when $y < H$. A typical projectile launched below $H$ with enough speed will: (1) rise through $y < H$ with wind acting, (2) travel above $H$ with no wind, then (3) descend back below $H$ with wind acting again before landing. Answer C correctly captures all three phases. During phases 1 and 3, you use $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; during phase 2, $m\ddot{x} = 0$, $m\ddot{y} = -mg$. At each boundary (crossing $y = H$ upward and downward), you match position and velocity to carry information forward, ultimately yielding the correct total range. A is wrong because it applies wind throughout the entire flight — including above $H$ — which violates the stated condition. The descent re-entering $y < H$ doesn't retroactively justify a constant $F_w$ everywhere. B is wrong because it ties the wind cutoff to $\dot{y} = 0$ (the apex), not to the altitude $y = H$. The apex and the moment of crossing $H$ are generally different instants. D is wrong because it correctly identifies the upward crossing of $H$ but then ignores the descent phase below $H$, missing the third phase where wind resumes. Your strategy: whenever a force has a conditional domain (position-dependent, speed-dependent, etc.), sketch the trajectory and list every interval where conditions change — each is a separate phase requiring its own equations of motion.
6. Which of the following correctly characterizes the nature of the motion in each coordinate direction and identifies whether the particle will remain bounded?
   Answer: The $x$-motion diverges exponentially away from $x = 0$ and is unbounded, while the $y$-motion is simple harmonic and bounded; the particle is unbounded because the $x$-force $F_x = 12x$ is a repulsive (unstable) force proportional to displacement. When you see a position-dependent force linear in displacement, your first instinct should be to check the sign of the proportionality constant — it determines whether the force is restoring (leading to oscillation) or repulsive (leading to exponential growth). Applying Newton's second law separately to each coordinate: $\ddot{x} = \frac{12x}{4} = 3x \quad \Rightarrow \quad \ddot{x} - 3x = 0$ $\ddot{y} = \frac{-8y}{4} = -2y \quad \Rightarrow \quad \ddot{y} + 2y = 0$ The $x$-equation has the form $\ddot{x} = +\omega^2 x$, which gives exponential solutions $x(t) = A e^{\sqrt{3}\,t} + B e^{-\sqrt{3}\,t}$. With $x(0)=1$ and $\dot{x}(0)=0$, both coefficients are nonzero, so $x(t)$ grows without bound. The $y$-equation has the form $\ddot{y} = -\omega^2 y$, which gives simple harmonic motion — bounded oscillation forever. This confirms answer C is correct. A gets the physics exactly backwards — it mistakenly calls the $x$-force "restoring" and the $y$-force "repulsive," when the signs say the opposite. B incorrectly assumes any force linear in displacement is harmonic. Linearity alone isn't enough; the sign determines oscillation versus exponential behavior. D contains a subtle trap: the two equations of motion are decoupled ($x$ and $y$ appear independently), so no energy transfer between directions is possible. The $y$-direction cannot "rescue" $x$, and calling the system "bounded overall" is simply wrong. Key tip: Always write out the equation $\ddot{q} + kq = 0$. If $k > 0$, it's SHM; if $k < 0$, it's exponential growth — sign is everything.

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