Elastic Collisions in One Dimension — Question 6

Section 9-9 · Physics pages 1322–1323

Question 6 A sequence of equal-mass impacts transmits velocity Three identical carts A, B and C lie in that order on a frictionless line with positive gaps. Initially A moves right at 6 m/s while B and C are at rest. Each collision is isolated, one dimensional and elastic. A B C 6 Schematic; labels, not drawing lengths, specify the data. (a) Find the velocities after A hits B and after B hits C. (b) Explain why no later collisions occur and calculate the total impulse received by B over the sequence.
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Question 6 - Solution A sequence of equal-mass impacts transmits velocity A B C 6 (a) Derivation and calculation An elastic collision of identical masses exchanges their velocities: (6, 0, 0) A−B −−−→(0, 6, 0) B−C −−−→(0, 0, 6) m/s. vA = 0, vB = 0, vC = 6 m/s . (b) Result and interpretation The two impulses on B are JB,1 = m(6 −0) = +6m, JB,2 = m(0 −6) = −6m. JB,net = 0 . B is accelerated and then stopped. Zero net impulse over both collisions does not mean zero impulse at each contact; the final velocities produce no further approach. Physical check All the initial momentum and kinetic energy end in C.

Question 6

A sequence of equal-mass impacts transmits velocity

Three identical carts A, B and C lie in that order on a frictionless line with positive gaps. Initially A moves right at 6m/s6\,\mathrm{m/s} while B and C are at rest. Each collision is isolated, one dimensional and elastic.

Diagram shown in the original SVG worksheet.

  1. Find the velocities after A hits B and after B hits C.

  2. Explain why no later collisions occur and calculate the total impulse received by B over the sequence.

Question 6 - Solution

A sequence of equal-mass impacts transmits velocity

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

An elastic collision of identical masses exchanges their velocities: (6,0,0)→A−B(0,6,0)→B−C(0,0,6)m/s.(6,0,0)\xrightarrow{A-B}(0,6,0)\xrightarrow{B-C}(0,0,6)\ \mathrm{m/s}. vA=0,vB=0,vC=6m/s.\boxed{v_A=0,\quad v_B=0,\quad v_C=6\ \mathrm{m/s}}.

(b) Result and interpretation

The two impulses on B are JB,1=m(6−0)=+6m,JB,2=m(0−6)=−6m.J_{B,1}=m(6-0)=+6m,\qquad J_{B,2}=m(0-6)=-6m. JB,net=0.\boxed{J_{B,\mathrm{net}}=0}. B is accelerated and then stopped. Zero net impulse over both collisions does not mean zero impulse at each contact; the final velocities produce no further approach.

Physical check

All the initial momentum and kinetic energy end in C.