Inelastic Collisions in One Dimension — Question 9

Section 9-8 · Physics pages 1308–1309

Question 9 A measured energy loss determines restitution Two 2 kg carts approach at +3 and −1 m/s. Their isolated head-on collision reduces kinetic energy by 6 J. They then separate. 2 2 +x Cart schematic; velocities or forces are specified in the text. (a) Find the separation speed and coefficient of restitution. (b) Find each final velocity and identify the branch rejected by the separation condition.
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Question 9 - Solution A measured energy loss determines restitution 2 2 +x (a) Derivation and calculation The center velocity and reduced mass are VC = 2(3) + 2(−1) 4 = 1 m/s, µ = 1 kg. The initial relative energy is 8 J, so after losing 6 J, 1 2µw2 f = 8 −6 = 2 ⇒wf = 2 m/s, e = 2/4 = 0.50 . (b) Result and interpretation Select the separating branch: v1 + v2 = 2VC = 2, v2 −v1 = 2. v1 = 0, v2 = 2 m/s . The alternative (2, 0) has the same energy and momentum but is still approaching. Physical check Total energy falls from 10 J to 4 J, exactly the specified 6 J loss.

Question 9

A measured energy loss determines restitution

Two 2kg2\,\mathrm{kg} carts approach at +3+3 and −1m/s-1\,\mathrm{m/s}. Their isolated head-on collision reduces kinetic energy by 6J6\,\mathrm J. They then separate.

Diagram shown in the original SVG worksheet.

  1. Find the separation speed and coefficient of restitution.

  2. Find each final velocity and identify the branch rejected by the separation condition.

Question 9 - Solution

A measured energy loss determines restitution

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The center velocity and reduced mass are VC=2(3)+2(−1)4=1m/s,μ=1kg.V_C=\frac{2(3)+2(-1)}4=1\ \mathrm{m/s},\qquad\mu=1\ \mathrm{kg}. The initial relative energy is 8 J, so after losing 6 J, 12μwf2=8−6=2⇒wf=2m/s,e=2/4=0.50.\tfrac12\mu w_f^2=8-6=2\Rightarrow w_f=2\ \mathrm{m/s},\qquad\boxed{e=2/4=0.50}.

(b) Result and interpretation

Select the separating branch: v1+v2=2VC=2,v2−v1=2.v_1+v_2=2V_C=2,\qquad v_2-v_1=2. v1=0,v2=2m/s.\boxed{v_1=0,\qquad v_2=2\ \mathrm{m/s}}. The alternative (2,0)(2,0) has the same energy and momentum but is still approaching.

Physical check

Total energy falls from 10 J to 4 J, exactly the specified 6 J loss.