Kinetic Energy — Question 6

Section 7-2 · Physics pages 882–883

Question 6 A squared-speed plot separates mass from a sensor offset A sensor reads R = 1 2mv2 + b, where b is an unknown constant offset in joules. The three exact measurements below concern the same particle. v2 (m2/s2) R (J) (4, 11) (16, 29) (36, 59) Three sensor readings constrain the slope and additive offset. (a) Find m and b. (b) Find the true kinetic energy at 4 m/s. (c) Predict R at rest and explain why forcing the graph through the origin would misestimate mass.
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Question 6 - Solution A squared-speed plot separates mass from a sensor offset v2 (m2/s2) R (J) (4, 11) (16, 29) (36, 59) (a) Use differences to eliminate the offset. The governing relation is R = sv2 + b, s = ∆R ∆(v2) = m 2 Eliminate the unknown offset by subtracting two readings: m 2 = 29 −11 16 −4 = 1.5 kg =⇒m = 3 kg . b = 11 −1 2(3)(4) = 5 J , R(36) = 1 2(3)(36) + 5 = 59 J. The third point independently agrees with the inferred model. (b) Remove the sensor offset. At v = 4 m/s, subtract the sensor offset: K = 1 2(3)(4)2 = 24 J , R = K + b = 29 J. (c) Evaluate zero speed. The physical kinetic energy vanishes at rest, but the reading does not: K(0) = 0, R(0) = b = 5 J. Forcing a zero intercept through the first point instead gives mfalse = 2R v2 = 2(11) 4 = 5.5 kg̸ = 3 kg. That fit assigns the offset to the mass.

Question 6

A squared-speed plot separates mass from a sensor offset

A sensor reads R=12mv2+bR=\tfrac12 m v^2+b, where b is an unknown constant offset in joules. The three exact measurements below concern the same particle.

Diagram shown in the original SVG worksheet.

  1. Find m and b.

  2. Find the true kinetic energy at 4 m/s.

  3. Predict R at rest and explain why forcing the graph through the origin would misestimate mass.

Question 6 - Solution

A squared-speed plot separates mass from a sensor offset

Diagram shown in the original SVG worksheet.

(a) Use differences to eliminate the offset.

The governing relation is R=sv2+b,s=ΔRΔ(v2)=m2\boxed{R=s v^2+b,\qquad s=\frac{\Delta R}{\Delta(v^2)}=\frac m2} Eliminate the unknown offset by subtracting two readings: m2=29−1116−4=1.5kg⇒m=3kg.\frac m2=\frac{29-11}{16-4}=1.5\ \mathrm{kg}\Longrightarrow\boxed{m=3\ \mathrm{kg}}. b=11−12(3)(4)=5J,R(36)=12(3)(36)+5=59J.\boxed{b=11-\tfrac12(3)(4)=5\ \mathrm J},\qquad R(36)=\tfrac12(3)(36)+5=59\ \mathrm J. The third point independently agrees with the inferred model.

(b) Remove the sensor offset.

At v=4v=4 m/s, subtract the sensor offset: K=12(3)(4)2=24J,R=K+b=29J.\boxed{K=\tfrac12(3)(4)^2=24\ \mathrm J},\qquad R=K+b=29\ \mathrm J.

(c) Evaluate zero speed.

The physical kinetic energy vanishes at rest, but the reading does not: K(0)=0,R(0)=b=5J.K(0)=0,\qquad R(0)=b=5\ \mathrm J. Forcing a zero intercept through the first point instead gives mfalse=2Rv2=2(11)4=5.5kg≠3kg.m_{\mathrm{false}}=\frac{2R}{v^2}=\frac{2(11)}4=5.5\ \mathrm{kg}\ne3\ \mathrm{kg}. That fit assigns the offset to the mass.