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Physics / Chapter 9: Center of Mass and Linear Momentum / Section 9-8

Inelastic Collisions in One Dimension — Question 6

Section 9-8 · Physics pages 1302–1303

Question 6 Two successive stationary captures partition the loss A 2 kg cart at 6 m/s captures a stationary 1 kg cart, then captures a second stationary 1 kg cart farther along the frictionless track. All captures are perfectly inelastic. 2 kg 1 kg 1 kg 6 m/s two successive captures The isolated train gains mass at each stationary pickup. (a) Find the speeds after each capture and the losses in each event. (b) Compare the final state with simultaneous capture of both carts.
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Question 6 - Solution Two successive stationary captures partition the loss 2 kg 1 kg 1 kg 6 m/s two successive captures (a) Derivation and calculation At the first capture, 2(6) = 3V1 ⇒V1 = 4 m/s , L1 = 36 −1 2(3)(4)2 = 12 J. At the second, 3(4) = 4V2 ⇒V2 = 3 m/s , L2 = 24 −1 2(4)(3)2 = 6 J. (b) Result and interpretation For simultaneous capture of both masses, 2(6) = 4V ⇒V = 3 m/s, L = 36 −1 2(4)(3)2 = 18 J = L1 + L2 . The intermediate energy history changes, but the initial and final states agree. Physical check The final momentum remains 12 kg m/s in either sequence.

Question 6

Two successive stationary captures partition the loss

A 2kg2\,\mathrm{kg} cart at 6m/s6\,\mathrm{m/s} captures a stationary 1kg1\,\mathrm{kg} cart, then captures a second stationary 1kg1\,\mathrm{kg} cart farther along the frictionless track. All captures are perfectly inelastic.

Diagram shown in the original SVG worksheet.

  1. Find the speeds after each capture and the losses in each event.

  2. Compare the final state with simultaneous capture of both carts.

Question 6 - Solution

Two successive stationary captures partition the loss

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

At the first capture, 2(6)=3V1⇒V1=4m/s,L1=36−12(3)(4)2=12J.2(6)=3V_1\Rightarrow\boxed{V_1=4\ \mathrm{m/s}},\qquad L_1=36-\tfrac12(3)(4)^2=12\ \mathrm J. At the second, 3(4)=4V2⇒V2=3m/s,L2=24−12(4)(3)2=6J.3(4)=4V_2\Rightarrow\boxed{V_2=3\ \mathrm{m/s}},\qquad L_2=24-\tfrac12(4)(3)^2=6\ \mathrm J.

(b) Result and interpretation

For simultaneous capture of both masses, 2(6)=4V⇒V=3m/s,2(6)=4V\Rightarrow V=3\ \mathrm{m/s}, L=36−12(4)(3)2=18J=L1+L2.\boxed{L=36-\tfrac12(4)(3)^2=18\ \mathrm J=L_1+L_2}. The intermediate energy history changes, but the initial and final states agree.

Physical check

The final momentum remains 12 kg m/s in either sequence.

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