Case Study: Air Track Collision – Practical Data Analysis | 案例分析：气垫导轨碰撞实验

📚 Case Study: Air Track Collision – Practical Data Analysis | 案例分析：气垫导轨碰撞实验 – 实战数据分析

In Year 12 CAIE Physics, being able to analyse experimental data is just as important as understanding the theory. This case study takes you through a complete practical investigation of a collision on an air track, from recording raw measurements to drawing conclusions about momentum, energy, forces, and errors. Follow every step and you will sharpen the exact skills needed for Paper 3 and data‑analysis questions.

在12年级CAIE物理中，分析实验数据的能力与理解理论知识同等重要。本篇案例将带你完整走一遍气垫导轨碰撞实验的探究过程——从记录原始数据到得出关于动量、能量、力以及误差的结论。紧跟每一步，你将精准磨炼Paper 3和数据分析题所需的技能。

1. The Scenario and Aims | 实验情景与目标

Two laboratory gliders, A and B, are placed on a linear air track. Glider B is initially at rest. Glider A is launched gently so that it collides with B. The gliders couple together on impact and move as a single combined object.

两辆实验室滑块A和B置于直线气垫导轨上。滑块B最初静止。轻推滑块A使其与B发生碰撞，碰撞后两滑块扣合在一起，作为一个整体继续运动。

The aim is to use photogate timing to investigate conservation of linear momentum, calculate the kinetic energy lost, estimate the average impact force, and evaluate the uncertainties in the experiment.

实验目标是通过光电门计时来探究线动量守恒，计算动能损失，估算平均碰撞力，并对实验的不确定度进行评估。

2. Equipment Setup and Raw Data | 装置与原始数据

The air track is fitted with two photogates. Gate 1 measures the speed of glider A just before the impact. Gate 2 measures the speed of the coupled gliders just after the collision, once they have travelled a short distance. Each glider carries a card of known width that interrupts the beam.

气垫导轨上装有两个光电门。光电门1测量滑块A撞击前的瞬时速度，光电门2测量两滑块扣合后运动一段短距离后的速度。每辆滑块都装有已知宽度的挡光片，用于遮挡光束。

Quantity	Value / Unit	量	数值/单位
Mass of glider A, m_A	0.220 kg	滑块A质量 m_A	0.220 kg
Mass of glider B, m_B	0.260 kg	滑块B质量 m_B	0.260 kg
Card width, d	0.0500 m	挡光片宽度 d	0.0500 m
Collision duration (from force sensor), Δt	0.048 s	碰撞持续时间（力传感器）Δt	0.048 s

The interruption times recorded by each photogate are shown below. Five repeat runs were carried out to allow averaging and error estimation.

每个光电门记录到的遮挡时间如下表所示。进行了五次重复测量以便求平均值并进行误差估算。

Run	t₁ / s (Gate 1, before collision)	t₂ / s (Gate 2, after collision)	实验次数	t₁ / s（光电门1，碰前）	t₂ / s（光电门2，碰后）
1	0.125	0.308	1	0.125	0.308
2	0.124	0.310	2	0.124	0.310
3	0.126	0.305	3	0.126	0.305
4	0.124	0.309	4	0.124	0.309
5	0.125	0.311	5	0.125	0.311

3. Processing the Timings into Velocities | 将时间数据换算为速度

The instantaneous speed of a glider is calculated using v = d / t, where d is the card width and t is the interruption time. We first work out the mean time for each gate.

滑块的瞬时速度由 v = d / t 计算，其中 d 为挡光片宽度，t 为遮挡时间。我们首先算出每个光电门所测时间的平均值。

Mean gate-1 time, t₁ = (0.125 + 0.124 + 0.126 + 0.124 + 0.125) / 5 = 0.1248 s.

光电门1平均时间 t₁ = (0.125 + 0.124 + 0.126 + 0.124 + 0.125) / 5 = 0.1248 s。

Mean gate-2 time, t₂ = (0.308 + 0.310 + 0.305 + 0.309 + 0.311) / 5 = 0.3086 s.

光电门2平均时间 t₂ = (0.308 + 0.310 + 0.305 + 0.309 + 0.311) / 5 = 0.3086 s。

Hence, the initial speed of glider A is u_A = d / t₁ = 0.0500 m / 0.1248 s = 0.4006 m/s ≈ 0.401 m/s.

因此，滑块A的初始速度 u_A = d / t₁ = 0.0500 m / 0.1248 s = 0.4006 m/s ≈ 0.401 m/s。

The final speed of the coupled pair is v_f = d / t₂ = 0.0500 m / 0.3086 s = 0.1620 m/s ≈ 0.162 m/s.

两滑块扣合后的末速度 v_f = d / t₂ = 0.0500 m / 0.3086 s = 0.1620 m/s ≈ 0.162 m/s。

We will quote the speeds to three significant figures, consistent with the precision of the measured times.

我们保留三位有效数字，以与测量时间的精度相匹配。

4. Testing Conservation of Linear Momentum | 检验线动量守恒

Total momentum before the collision should equal total momentum after the collision if the system is isolated. Glider B is stationary before the impact, so the initial momentum is simply p_before = m_A u_A.

如果系统无外力，碰撞前的总动量应等于碰撞后的总动量。滑块B在碰撞前静止，因此初始动量仅为 p_before = m_A u_A。

p_before = 0.220 kg × 0.401 m/s = 0.0882 kg m/s (to 3 s.f.).

p_before = 0.220 kg × 0.401 m/s = 0.0882 kg·m/s（三位有效数字）。

After the collision the two gliders move together, so the final momentum is p_after = (m_A + m_B) × v_f = (0.220 kg + 0.260 kg) × 0.162 m/s = 0.480 kg × 0.162 m/s = 0.0778 kg m/s.

碰撞后两滑块一起运动，末动量为 p_after = (m_A + m_B) × v_f = (0.220 kg + 0.260 kg) × 0.162 m/s = 0.480 kg × 0.162 m/s = 0.0778 kg·m/s。

The percentage difference between after and before momentum is [(0.0778 – 0.0882) / 0.0882] × 100% = -11.8%. Momentum appears to be lower after the collision, suggesting a loss that needs investigating.

碰撞后与碰撞前动量的百分比差异为 [(0.0778 – 0.0882) / 0.0882] × 100% = -11.8%。碰撞后动量似乎减少了，这一亏损需要进一步探究。

5. Accounting for the Momentum ‘Loss’ | 解释动量的“损失”

In a perfectly isolated air‑track collision, momentum should be conserved within experimental uncertainty. The apparent 12% drop implies systematic effects are at work.

在一个理想无摩擦的气垫导轨碰撞中，动量应在实验误差范围内守恒。约12%的明显下降意味着存在系统效应。

Possible causes include a small frictional force on the track, air resistance on the gliders, and the finite precision of the photogates. Since the gliders are light, even a tiny external force integrated over time can produce a noticeable momentum change.

可能的原因包括导轨上存在微小摩擦力、滑块受到的空气阻力，以及光电门的有限精度。由于滑块很轻，即便微小的外力经过时间积分也能产生可观的动量变化。

However, the most likely culprit is that the card width was measured with a vernier caliper but the beam of the photogate is not perfectly aligned with the card’s edge, introducing a systematic timing error. We will quantify this in the error analysis.

不过，最大的可能性在于：虽然用游标卡尺测量了挡光片宽度，但光电门光束并未与挡光片边缘完美对齐，从而引入了系统性的计时误差。我们将在误差分析中对此进行量化。

6. Kinetic Energy and the Inelastic Collision | 动能与非弹性碰撞

Because the gliders couple together, the collision is completely inelastic. Kinetic energy is not conserved; some of it is dissipated as heat and sound during the coupling mechanism.

由于两滑块扣合在一起，该碰撞属于完全非弹性碰撞。动能不守恒，部分动能会在扣合过程中转化为热量和声音。

Initial kinetic energy: E_k,before = ½ m_A u_A² = ½ × 0.220 kg × (0.401 m/s)² = 0.110 × 0.1608 = 0.0177 J.

初始动能：E_k,before = ½ m_A u_A² = ½ × 0.220 kg × (0.401 m/s)² = 0.110 × 0.1608 = 0.0177 J。

Final kinetic energy: E_k,after = ½ (m_A + m_B) v_f² = ½ × 0.480 kg × (0.162 m/s)² = 0.240 × 0.02624 = 0.00630 J.

末动能：E_k,after = ½ (m_A + m_B) v_f² = ½ × 0.480 kg × (0.162 m/s)² = 0.240 × 0.02624 = 0.00630 J。

Energy dissipated = E_k,before – E_k,after = 0.0177 J – 0.00630 J = 0.0114 J. This represents about 64% of the original kinetic energy — typical for a soft coupling.

耗散能量 = 0.0177 J – 0.00630 J = 0.0114 J，约占初始动能的64%——这对于软性扣合来说是典型的数值。

7. Estimating the Average Impact Force | 估算平均撞击力

The impulse–momentum theorem states that the impulse exerted on glider B equals its change in momentum: F_avg Δt = Δp. Glider B accelerates from rest to the common final speed, so its change in momentum is Δp_B = m_B v_f – 0.

冲量 – 动量定理表明，施加于滑块B的冲量等于其动量变化量：F_avg Δt = Δp。滑块B从静止加速到共同末速度，故其动量变化量为 Δp_B = m_B v_f – 0。

Δp_B = 0.260 kg × 0.162 m/s = 0.0421 kg m/s. With a collision duration Δt = 0.048 s, the average force on B is F_avg = Δp_B / Δt = 0.0421 / 0.048 = 0.877 N (approx 0.88 N).

Δp_B = 0.260 kg × 0.162 m/s = 0.0421 kg·m/s。已知碰撞持续时间 Δt = 0.048 s，则作用于B的平均力 F_avg = Δp_B / Δt = 0.0421 / 0.048 = 0.877 N（约 0.88 N）。

By Newton’s third law, glider A experiences a force of equal magnitude in the opposite direction. This is confirmed by calculating the impulse on A: Δp_A = m_A v_f – m_A u_A = 0.220 × (0.162 – 0.401) = -0.0526 kg m/s, giving an average force of -1.10 N. The slight difference from 0.88 N highlights the impact of the systematic errors we noted.

根据牛顿第三定律，滑块A受到大小相等、方向相反的力。计算A的冲量可加以验证：Δp_A = m_A v_f – m_A u_A = 0.220 × (0.162 – 0.401) = -0.0526 kg·m/s，所得平均力为 -1.10 N。与 0.88 N 的微小差异再次突显了前述系统误差的影响。

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Case Study: Air Track Collision – Practical Data Analysis | 案例分析：气垫导轨碰撞实验 – 实战数据分析