Inelastic Collisions in One Dimension — Question 1

Section 9-8 · Physics pages 1292–1293

Question 1 A head-on capture can end at rest with no energy left in translation A 3 kg cart moving at +4 m/s sticks to a 2 kg cart moving at −6 m/s. The track is frictionless. 3 2 +x Cart schematic; velocities or forces are specified in the text. (a) Find their common velocity and kinetic-energy loss. (b) Find the impulse on each cart. Explain how nonzero impulses coexist with zero system impulse.
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Question 1 - Solution A head-on capture can end at rest with no energy left in translation 3 2 +x (a) Derivation and calculation For the isolated pair, Pi = 3(4) + 2(−6) = 0 = (3 + 2)V ⇒V = 0 . Ki = 1 2(3)(4)2 + 1 2(2)(6)2 = 60 J, Kf = 0. Kloss = 60 J . (b) Result and interpretation The individual impulses follow from their momentum changes: J1 = 3(0 −4) = −12 N s , J2 = 2(0 + 6) = +12 N s . Their sum is zero, although each object experiences a nonzero impulse. Physical check Zero system momentum does not imply zero initial kinetic energy.

Question 1

A head-on capture can end at rest with no energy left in translation

A 3kg3\,\mathrm{kg} cart moving at +4m/s+4\,\mathrm{m/s} sticks to a 2kg2\,\mathrm{kg} cart moving at −6m/s-6\,\mathrm{m/s}. The track is frictionless.

Diagram shown in the original SVG worksheet.

  1. Find their common velocity and kinetic-energy loss.

  2. Find the impulse on each cart. Explain how nonzero impulses coexist with zero system impulse.

Question 1 - Solution

A head-on capture can end at rest with no energy left in translation

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For the isolated pair, Pi=3(4)+2(−6)=0=(3+2)V⇒V=0.P_i=3(4)+2(-6)=0=(3+2)V\Rightarrow\boxed{V=0}. Ki=12(3)(4)2+12(2)(6)2=60J,Kf=0.K_i=\tfrac12(3)(4)^2+\tfrac12(2)(6)^2=60\ \mathrm J,\qquad K_f=0. Kloss=60J.\boxed{K_{\mathrm{loss}}=60\ \mathrm J}.

(b) Result and interpretation

The individual impulses follow from their momentum changes: J1=3(0−4)=−12Ns,J2=2(0+6)=+12Ns.\boxed{J_1=3(0-4)=-12\ \mathrm{N\,s}},\qquad\boxed{J_2=2(0+6)=+12\ \mathrm{N\,s}}. Their sum is zero, although each object experiences a nonzero impulse.

Physical check

Zero system momentum does not imply zero initial kinetic energy.