Collision and Impulse — Question 8

Question 8

Calibration bias contaminates the impulse area

A force sensor reads $F_{\mathrm{read}}=F_{\mathrm{true}}+5.0\,\mathrm N$ during a $0.20\,\mathrm s$ collision. Integrating the displayed trace gives $3.0\,\mathrm{N\,s}$. A $0.40\,\mathrm{kg}$ object initially moves at $-2.0\,\mathrm{m/s}$; this is its only force along $x$.

1. Find the corrected impulse and final velocity.

2. Find the velocity error caused by using the uncorrected trace.

Question 8 - Solution

Calibration bias contaminates the impulse area

(a) Derivation and calculation

Subtract the bias area from the measured integral: $$J_{\mathrm{bias}}=5.0(0.20)=1.0\ \mathrm{N\,s},\qquad \boxed{J_{ \mathrm{true}}=3.0-1.0=2.0\ \mathrm{N\,s}}.$$ $$\boxed{v_f=v_i+J_{\mathrm{true}}/m=-2.0+2.0/0.40=3.0\ \mathrm{m/s}}.$$

(b) Result and interpretation

The uncorrected result and its error are $$v_{\mathrm{read}}=-2.0+3.0/0.40=5.5\ \mathrm{m/s},\qquad \boxed{\delta v=J_{\mathrm{bias}}/m=1.0/0.40=2.5\ \mathrm{m/s}}.$$ Even a constant sensor offset accumulates a finite momentum error.

Physical check

The corrected momentum changes from $-0.8$ to $1.2$ kg m/s, a 2 N s increase.