Self-check: 203-SN1-RE, Mechanics

b) With the product rule: $\begin{aligned}&h'(x) \\ &\quad = (2x)(x^{3}-2x)+(x^{2}+1)(3x^{2}-2) \\ &\quad = 2x^{4}-4x^{2}+3x^{4}-2x^{2}+3x^{2}-2 \\ &\quad = 5x^{4}-3x^{2}-2\end{aligned}$ Check by expanding first: $h(x)=x^{5}-x^{3}-2x$, so $h'=5x^{4}-3x^{2}-2$ ✓. Conclusion: both routes give the same result, but expanding first is only possible for polynomials; the product rule works on any product ($xe^{x}$, $x\sin x$). Pitfall: believing that $(fg)'=f'g'$. Here $f'g'=(2x)(3x^{2}-2)=6x^{3}-4x$, which has nothing to do with it.

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a) $f(x)=\sqrt{x}$, $f'(x)=\frac{1}{2\sqrt{x}}$, so $f(4)=2$ and $f'(4)=\frac{1}{4}$. The linearization is $L(x)=2+\frac{1}{4}(x-4)$. Estimate: $\sqrt{4.1}\approx L(4.1)=2+\frac{0.1}{4}=2.025$. Exact value: $\sqrt{4.1}\approx 2.0248457$. The absolute error is about $0.000154$, a relative error of $0.0076$ percent. For mental arithmetic, that is remarkable: three correct decimals with no tool at all.

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a) The smallest division is $1$ mm, so $\Delta L=0.5$ mm $=0.05$ cm. Each end of the object is read to within half a division.

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b) The area under the curve is a trapezoid, which we cut up: triangle $\frac{1}{2}(8)(20)=80$ m, rectangle $(20)(20)=400$ m, triangle $\frac{1}{2}(8)(20)=80$ m. Total distance: $80+400+80=560$ m.

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a) “$20^{\circ}$ west of north” reads in order: you start from NORTH and turn $20^{\circ}$ toward the west. The angle is therefore measured from the $y$-axis, and it is the SINE that gives the $x$-component: $\Delta x_{2}=-3.00\sin 20.0^{\circ}=-1.03$ km (toward the west, so negative) and $\Delta y_{2}=3.00\cos 20.0^{\circ}=2.82$ km. The classic trap is to write $3\cos 20^{\circ}$ for $x$ out of reflex: you would get a leg pointing almost due east, which the figure contradicts at once. Safe rule: the cosine goes on the axis you START FROM when measuring the angle.

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a) Threshold of slipping: $f_{s,max}=\mu_{s}N=\mu_{s}mg=(0.50)(20)(9.8)=98$ N. Since 90 N $<$ 98 N, the crate does not move. The friction then equals exactly 90 N (not 98 N!): static friction adjusts itself to cancel the net force, and reaches its maximum only at the threshold of slipping.

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b) Here we must conserve energy, not balance forces. The block starts from rest and stops at the lowest point, so $\Delta K=0$: all the gravitational potential energy lost has been stored in the spring. $mgx_{max}=\frac{1}{2}kx_{max}^{2}$, whence $x_{max}=\frac{2mg}{k}=0.157$ m, or $15.7$ cm. That is EXACTLY double the equilibrium extension, and the result is general: $x_{max}=2x_{eq}$, whatever the mass and the stiffness. A load released suddenly stresses a structure twice as much as a load lowered gently, which is a basic principle of engineering design.

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b) For a projectile hitting a stationary target, solving the system gives $v_{1}'=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}=\frac{1-3}{4}(4)=-2$ m/s and $v_{2}'=\frac{2m_{1}}{m_{1}+m_{2}}v_{1}=\frac{2}{4}(4)=2$ m/s. The light cart BOUNCES BACK ($v_{1}'<0$): it hits something heavier than itself; the heavy cart moves forward at 2 m/s.

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e) First error: the theorem $I=I_{cm}+Md^{2}$ MUST start from the centre of mass. Let us look for the moment of inertia of the rod alone about an axis a quarter of the way along it, $0.200$ m from the centre. The correct answer is $0.0640+1.20(0.200)^{2}=0.112$ kg$\cdot$m$^{2}$. Starting wrongly from the end, you would write $0.256+1.20(0.200)^{2}=0.304$ kg$\cdot$m$^{2}$, almost triple. The theorem is not a general rule for shifting an axis: it is a relationship that singles out the centre of mass, and it follows that $I_{cm}$ is always the SMALLEST of the moments of inertia for a given direction of axis. Second error: the same rod of $1.20$ kg and $0.800$ m has $0.0640$ kg$\cdot$m$^{2}$ about its centre, $0.256$ kg$\cdot$m$^{2}$ about one end, and practically zero about its own long axis. Speaking of the moment of inertia of a body without naming the axis is meaningless, whereas speaking of its mass always makes sense.

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a) Rectangle 1, the base: area $A_{1}=2.00\times 0.50=1.00$ m², centre at $\left(1.00,\ 0.25\right)$. Rectangle 2, the upright: area $A_{2}=0.50\times 1.80=0.90$ m², centre at $\left(0.25,\ 0.50+0.90\right)=\left(0.25,\ 1.40\right)$, since it rests on the first.

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c) We want $\dfrac{GM}{r^{2}}=\dfrac{1}{2}\dfrac{GM}{R^{2}}$, so $r^{2}=2R^{2}$ and $r=R\sqrt{2}=9.010\times 10^{6}$ m. The altitude is $r-R=2.639\times 10^{6}$ m, about 2640 km. You have to rise by more than a third of an Earth radius for gravity to drop by half: it decreases much more slowly than intuition suggests.

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