Ballistic Pendulum with a Tangential Spring
A 0.045 kg steel ball is fired horizontally into a 0.180 kg pendulum bob at rest, embedding in it. The bob hangs from a light string of length 0.60 m, and the combined mass swings up. A student measures the maximum swing angle as 22.0 degrees using a protractor taped to the apparatus; the protractor reading is taken by eye and the string has visible slack at the pivot. The student predicts the firing speed using m v = (M+m) V, then (M+m) g L (1-cos theta) = (1/2)(M+m) V^2, obtaining v. A second trial with a stiffer spring gives a larger angle but the student reports the same speed, claiming momentum is conserved in both cases.
1. Identify the two most significant systematic uncertainties in this measurement and explain how each biases the inferred firing speed.
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Slack at the pivot. If the string is not taut during the swing, the effective pendulum length is shorter than 0.60 m and the pivot itself can translate. A shorter L at fixed angle gives a smaller height rise, so the inferred V and hence v are too small. The bias grows with angle and is not random.
Protractor reading by eye. Reading to about ±2 degrees at 22 degrees is roughly a 10% angle uncertainty; since height goes as (1-cos theta), the fractional error in height is amplified: d(1-cos theta)/(1-cos theta) = sin(theta)/(1-cos theta) dtheta ≈ (0.375/0.0728)(0.035) ≈ 0.18, so ~18% in height and ~9% in speed. This is random, not systematic, but it dominates the error budget.
A third issue is energy loss to deformation, but that is small for a steel-on-steel embed and is the standard approximation, not a dominant error here.
2. Evaluate the student's claim that momentum conservation makes the two trials equivalent.
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The claim confuses which conservation law applies in which interval.
- Collision (very short): momentum is conserved because external impulses (gravity, string tension) are negligible over the collision time. This is why m v = (M+m)V is valid.
- Swing (after collision): momentum is not conserved because the pivot/string exerts an external horizontal force; mechanical energy is conserved instead (minus small losses).
So the two trials are not equivalent in momentum terms: a stiffer spring means a larger v, a larger V, a larger angle, and a larger external impulse delivered by the pivot during the swing. Reporting the same speed for both trials contradicts the measured angles and is wrong. The correct conclusion is that v scales with the measured angle via v = (M+m)/m * sqrt(2 g L (1-cos theta)).
3. The student reports v = 6.0 m/s. Using the stated numbers, is this plausible? Show the calculation and state whether the reported value is consistent.
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Compute V from the energy step: V = sqrt(2 g L (1-cos theta)) = sqrt(2 * 9.8 * 0.60 * (1 - cos 22°)). cos 22° = 0.9272, so 1-cos = 0.0728. Then 29.80.60 = 11.76; times 0.0728 = 0.856; sqrt = 0.925 m/s.
Then v = (M+m)/m * V = (0.225/0.045) * 0.925 = 5.0 * 0.925 = 4.63 m/s.
The reported 6.0 m/s is about 30% higher than the value implied by the stated angle. Given the ~9% random error from the protractor and the downward bias from slack, 6.0 m/s is not consistent with the measurement. The student likely used an unmeasured or assumed angle, or omitted the mass ratio correctly. The value should be reported as about 4.6 m/s with an uncertainty of roughly ±0.4 m/s.