Physics

Physics · mechanics

Problem

a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum pp. i. At what height HH above the table is the impulse pp required to topple the domino smallest? ii. What is the minimum value of pp to topple the domino? Given constants: g=9.8 N/kgg=9.8 \mathrm{~N} / \mathrm{kg} k=1/4πϵ0=8.99×109 N⋅m2/C2k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} c=3.00×108 m/sc=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} NA=6.02×1023( mol)−1N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) 1eV=1.602×10−19 J1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} me=9.109×10−31 kg=0.511MeV/c2m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} mp=1.673×10−27 kg=938MeV/c2m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2} sin⁡θ≈θ−16θ3\sin \theta \approx \theta-\frac{1}{6} \theta^{3} for ∣θ∣≪1|\theta| \ll 1 G=6.67×10−11 N⋅m2/kg2km=μ0/4π=10−7 T⋅m/AkB=1.38×10−23 J/KR=NAkB=8.31 J/(mol⋅K)e=1.602×10−19Ch=6.63×10−34 J⋅s=4.14×10−15eV⋅s(1+x)n≈1+nx for ∣x∣≪1ln⁡(1+x)≈x for ∣x∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪1\begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned}  velocity parameter β=vc Lorentz factor γ=11−β2 relativistic momentum p=γβmc relativistic energy E=γmc2 relativistic doppler shift ff0=1−β1+β\begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} Extra Supplementary Reading Materials: Suppose a domino stands upright on a table. It has height hh, thickness tt, width ww (as shown below), and mass mm. The domino is free to rotate about its edges, but will not slide across the table. <image_1>

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