Collisions, work and oscillator energy
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| external impulse | 外力冲量 | wài lì chōng liàng |
| turning point/ˈtɜːnɪŋ pɔɪnt/ | 转折点 | zhuǎn zhé diǎn |
A decision before an answer
- Two carts approach from different directions. After they stick, their momentum can be conserved while their kinetic energy falls.
- Your goal: Compute vector momentum and energy loss in sticking collisions.
Read the relationship
- Choose an isolated system during the short collision. With negligible external impulse, 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.
- Compare fixed-force springs and use work–energy conditions.
For the two carts in the worked example, the common velocity is:
Divide the vector momentum (6,3) by total mass 3, component by component.
Use the defining rule
- 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.
- Derive small-oscillation periods and oscillator energies.
Two springs with k1=3k2 hold the same force. The ratio U1/U2 is:
At fixed force U=F²/(2k), hence the inverse stiffness ratio.
Check 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.
- Derive small-oscillation periods and oscillator energies.
A 2 kg cart at (3,0) m/s sticks to a 1 kg cart at (0,3) m/s. Momentum is (6,3) kg·m/s, so final velocity is (2,1) m/s. Initial K=9+4.5=13.5 J; final K=½·3·5=7.5 J. Internal-energy increase is 6 J. Separately, a 0.5 kg oscillator has equilibrium speed 0.2 m/s: its total energy is 0.01 J, or 10 mJ.
A small-angle pendulum length is multiplied by 9 at fixed g. Its period is multiplied by ____.
T scales as sqrt(L).
Apply the task format
- For an undamped harmonic oscillator, total energy is ½m v²+½kx²=½kA². At equilibrium all its energy is kinetic; at a turning point 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.
- Derive small-oscillation periods and oscillator energies.
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.
Which answer fits this case?
Compute vector momentum and energy loss in sticking collisions
At equilibrium an undamped harmonic oscillator can have nonzero kinetic energy and zero spring potential energy.
Displacement is zero, so spring energy is zero; speed is maximal.
Keep the distinctions
- external impulse 外力冲量 — Time integral of the net force from outside the chosen system.
- turning point 转折点 — An extreme oscillator position where its instantaneous speed is zero.
- 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.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.