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CM.1 · Collisions, work and oscillator energy

GRE · GRE Subject Test · GRE 物理 · 知识点 10

训练
10.1

碰撞、功与振荡器能量

Two carts approach from different directions. After they stick, their momentum can be conserved while their kinetic energy falls.

Prerequisites: 1.

  • Compute vector momentum and energy loss in sticking collisions
  • Compare fixed-force springs and use work–energy conditions
  • Derive small-oscillation periods and oscillator energies
10.2

Choose the system and model

Choose an isolated system during the short collision. With negligible external impulse 外力冲量, conserve total momentum separately in every Cartesian direction. If masses m1 and m2 stick, their common velocity is (m1 v1+m2 v2)/(m1+m2). Find the magnitude only after adding vectors. Initial kinetic energy is the sum of each ½m|v|²; final kinetic energy uses the total mass and the common speed. Their difference becomes internal energy or deformation, not missing momentum. A perfectly inelastic collision means sticking, not final rest.

词汇 训练
English 中文 拼音
external impulse/ekˈstɜːnl ˈɪmpʌls/ 外力冲量 wài lì chōng liàng
10.3

Use the governing relation

The work–energy theorem ΔK=∫F·dr uses net force along displacement. For a constant parallel force over distance s, ΔK=Fs. Speed doubling quadruples kinetic energy. Conservation of mechanical energy is a separate statement requiring no unaccounted nonconservative work. At a spring displacement x, U=½kx². Under the same applied force F, equilibrium extension is F/k and stored energy is F²/(2k): a stiffer spring stretches less and stores less energy. Under the same extension, it stores more. Identify which condition is held fixed.

10.4

Apply the conditions

Near a stable equilibrium x0, expand U≈U(x0)+½U″(x0)(x−x0)²; the effective stiffness is positive U″(x0). The motion obeys m xddot=−k_eff(x−x0), with ω=sqrt(k_eff/m) and period 2π/ω. A simple pendulum obeys θddot+(g/L)sinθ=0. For small angles sinθ≈θ, so T=2πsqrt(L/g). This approximation explains period–length scaling; a large amplitude requires a correction, and the mass cancels for an ideal pendulum.

10.5

Check the conclusion

For an undamped harmonic oscillator, total energy is ½m v²+½kx²=½kA². At equilibrium all its energy is kinetic; at a turning point 转折点 its speed is zero and all its energy is potential. Angular frequency is not ordinary frequency: ω=2πf. If the same oscillator has equilibrium speed vmax, A=vmax/ω. Distinguish the instant at which speed is measured from the release displacement, and keep SI units when a millijoule answer is requested.

词汇 训练
English 中文 拼音
turning point/ˈtɜːnɪŋ pɔɪnt/ 转折点 zhuǎn zhé diǎn
10.6

Worked method

Two carts stick with negligible external impulse. Add momentum vectors 动量矢量, then divide by total mass. Known: m1 = 2 kg, v1 = (3,0) m/s; m2 = 1 kg, v2 = (0,3) m/s.

$$\mathbf v_f=\frac{m_1\mathbf v_1+m_2\mathbf v_2}{m_1+m_2} =\frac{(2\,\mathrm{kg})(3,0)\,\mathrm{m/s}+(1\,\mathrm{kg})(0,3)\,\mathrm{m/s}}{3\,\mathrm{kg}} =(2,1)\,\mathrm{m/s}.$$
$$K_i=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2 =\tfrac12(2\,\mathrm{kg})(3\,\mathrm{m/s})^2+\tfrac12(1\,\mathrm{kg})(3\,\mathrm{m/s})^2=13.5\,\mathrm J.$$
$$K_f=\tfrac12(m_1+m_2)|\mathbf v_f|^2=\tfrac12(3\,\mathrm{kg})(5\,\mathrm{m^2/s^2})=7.5\,\mathrm J.$$
The 6 J difference becomes internal energy. Kinetic energy is not conserved in sticking.

Collisions, work and oscillator energy: GRE original diagram
Collisions, work and oscillator energy: original GRE teaching diagram.
词汇 训练
English 中文 拼音
momentum vectors 动量矢量 dòng liàng shǐ liàng
10.7

Check conditions and vocabulary

Do not add incoming speed magnitudes as momenta, or conserve kinetic energy in a sticking collision. Fixed-force and fixed-extension spring comparisons have opposite answers.

external impulse: Time integral of the net force from outside the chosen system.

turning point: An extreme oscillator position where its instantaneous speed is zero.

该知识点的互动课程

逐步完成,配合即时检查练习。

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