Collisions, work and oscillator energy
GRE Subject Test · GRE Physics Topic 10 6:31 English narration · English + 中文 subtitles burned in
Chapters
Transcript
Two carts approach from different directions and stick.
两辆小车从不同方向驶来并粘在一起。
Their total momentum is conserved, but their kinetic energy falls.
它们的总动量守恒，但动能减少。
This topic computes vector momentum and the energy lost in sticking collisions, compares springs under a fixed force, applies work and energy, and derives small-oscillation periods and oscillator energies.
本主题计算粘合碰撞中的矢量动量和能量损失，比较固定力作用下的弹簧，应用功能关系，并推导小振动周期和振子能量。
Choose an isolated system during the short collision.
在短暂碰撞期间选择一个孤立系统。
With negligible external impulse, conserve total momentum separately in every Cartesian direction.
外力冲量可忽略时，在每个笛卡尔方向上分别守恒总动量。
If masses m 1 and m 2 stick, their common velocity is m 1 v 1 plus m 2 v 2, all over m 1 plus m 2.
若质量 m1 和 m2 粘在一起，其共同速度等于 m1 v1 加 m2 v2，再除以 m1 加 m2。
Find the magnitude only after adding vectors.
只有在矢量相加之后才求大小。
Initial kinetic energy is the sum of each one half m v squared; final kinetic energy uses the total mass and the common speed.
初始动能是各物体二分之一 m v 平方之和；末动能使用总质量和共同速率计算。
Their difference becomes internal energy or deformation, not missing momentum.
两者之差转化为内能或形变，而不是丢失的动量。
A perfectly inelastic collision means sticking, not final rest.
完全非弹性碰撞意味着粘在一起，而不是最终静止。
The work-energy theorem, delta K equals the integral of F dot dr, uses net force along displacement.
功能定理，delta K 等于 F 点乘 dr 的积分，使用沿位移方向的合力。
For a constant parallel force over distance s, delta K equals F times s.
对于在距离 s 上恒定的平行力，delta K 等于 F 乘以 s。
Speed doubling quadruples kinetic energy.
速度加倍会使动能变为四倍。
Conservation of mechanical energy is a separate statement requiring no unaccounted nonconservative work.
机械能守恒是一个独立的陈述，要求没有未计入的非保守力做功。
At a spring displacement x, U equals one half k x squared.
在弹簧位移为 x 时，U 等于二分之一 k x 平方。
Under the same applied force F, equilibrium extension is F divided by k and stored energy is F squared divided by 2k.
在相同的施加力 F 下，平衡伸长量为 F 除以 k，储存的能量为 F 平方除以 2k。
A stiffer spring stretches less and stores less energy.
较硬的弹簧伸长量较小且储存能量较少。
Under the same extension, it stores more.
在相同的伸长量下，它储存更多能量。
Identify which condition is held fixed.
请确定哪个条件保持不变。
Near a stable equilibrium x sub zero, expand U approximately equal to U of x sub zero plus one half U double prime of x sub zero times the quantity x minus x sub zero squared.
在稳定平衡点 x₀ 附近，将 U 近似展开为 U(x₀) + ½U''(x₀)(x − x₀)²。
The effective stiffness is positive U double prime of x sub zero.
有效刚度为正，即 U''(x₀) > 0。
The motion obeys m x double dot equals negative k sub eff times the quantity x minus x sub zero, with omega equals the square root of k sub eff divided by m and period equals 2 pi divided by omega.
运动方程满足 mẍ = −k_eff(x − x₀)，其中 ω = √(k_eff/m)，周期为 2π/ω。
A simple pendulum obeys theta double dot plus g over L times sin of theta equals zero.
单摆满足 θ̈ + (g/L)sinθ = 0。
For small angles, sin of theta is approximately equal to theta, so T equals 2 pi times the square root of L divided by g.
在小角度下，sinθ ≈ θ，因此 T = 2π√(L/g)。
This approximation explains period length scaling. A large amplitude requires a correction, and the mass cancels for an ideal pendulum.
该近似解释了周期与摆长的比例关系；大振幅需要修正，且理想单摆的周期与质量无关。
For an undamped harmonic oscillator, the total energy is one half m v squared plus one half k x squared, which equals one half k A squared.
对于无阻尼简谐振子，总能量为二分之一 m v 平方加二分之一 k x 平方，等于二分之一 k A 平方。
At equilibrium all its energy is kinetic; at a turning point its speed is zero and all its energy is potential.
在平衡位置，全部能量为动能；在转折点，速度为零，全部能量为势能。
Angular frequency is not ordinary frequency; omega equals 2 pi f.
角频率不是普通频率：ω 等于 2π 乘以 f。
If the same oscillator has equilibrium speed v max, then A equals v max over omega.
若同一振子在平衡位置的速度为 v max，则 A 等于 v max 除以 ω。
Distinguish the instant at which speed is measured from the release displacement, and keep SI units when a millijoule answer is requested.
要区分测量速度的时刻与释放时的位移；题目要求以毫焦作答时，计算中保持国际单位制。
Two carts stick with negligible external impulse.
两辆小车碰撞后粘在一起，外力冲量可忽略不计。
Add momentum vectors, then divide by total mass.
将动量矢量相加，再除以总质量。
Known: m sub 1 equals 2 kg, v sub 1 equals the vector 3 comma 0 metres per second; m sub 2 equals 1 kg, v sub 2 equals the vector 0 comma 3 metres per second.
已知：m₁ = 2 kg，v₁ = (3, 0) m/s；m₂ = 1 kg，v₂ = (0, 3) m/s。
Add the momentum vectors and divide by the total mass: 2 kilograms times 3 comma 0, plus 1 kilogram times 0 comma 3, all over 3 kilograms, gives a final velocity of 2 comma 1 metres per second.
把动量矢量相加再除以总质量：2 千克乘 (3, 0) 加 1 千克乘 (0, 3)，再除以 3 千克，得到末速度 (2, 1) 米每秒。
The initial kinetic energy is one half times 2 times 3 squared, plus one half times 1 times 3 squared: 9 plus 4.5, which is 13.5 joules.
初动能等于二分之一乘 2 乘 3 的平方，加上二分之一乘 1 乘 3 的平方：9 加 4.5，等于 13.5 焦耳。
Afterwards, the 3 kilograms move at speed root 5, so the kinetic energy is one half times 3 times 5: 7.5 joules.
碰撞后，3 千克的整体以根号 5 的速率运动，所以动能等于二分之一乘 3 乘 5，即 7.5 焦耳。
The 6 J difference becomes internal energy.
6 J 的差值转化为内能。
Kinetic energy is not conserved in sticking.
在粘连过程中，动能不守恒。
Two equal masses approach with speed u along directions separated by angle theta and stick.
两个等质量物体以速率 u 沿夹角为 θ 的方向相向运动并粘在一起。
Derive final speed and the fraction of initial kinetic energy lost.
推导最终速率及初始动能损失的比例。
Check parallel, perpendicular and opposite incoming directions.
验证平行、垂直和反向入射情形。
Five marks.
五分。
The sum of equal velocity vectors has magnitude two u, absolute value of cosine of theta over two.
等速矢量的和的大小为2u乘以cos(θ/2)的绝对值。
Dividing momentum by total mass gives v sub f equals u, absolute value of cosine of theta over two.
动量除以总质量得到v_f等于u乘以cos(θ/2)的绝对值。
K sub i equals m u squared and K sub f equals m v sub f squared, so the lost fraction is sine squared of theta over two.
K_i等于mu平方，K_f等于mv_f平方，因此损失分数为sin²(θ/2)。
The three limits are zero, one half, and one.
三个极限值分别为0、1/2和1。
Award vector sum, speed, energy relation, lost fraction, and limits.
评分点包括矢量求和、速度、能量关系、损失分数及极限值。
The formula uses zero less than or equal to theta less than or equal to pi between directions.
该公式中方向间的夹角θ满足0 ≤ θ ≤ π。
Do not add incoming speed magnitudes as momenta, or conserve kinetic energy in a sticking collision.
不要将入射速率大小当作动量相加，也不要在完全非弹性碰撞中假设动能守恒。
Fixed-force and fixed-extension spring comparisons have opposite answers.
恒力与恒形变弹簧的比较结果截然相反。
external impulse: Time integral of the net force from outside the chosen system.
外力冲量：来自选定系统之外的合外力的时间积分。
turning point: An extreme oscillator position where its instantaneous speed is zero.
转折点：振荡器的一个极端位置，其瞬时速度为零。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆其含义，然后改变原始传递中的一个参数并检验极限情况。