Inelastic Collisions in One Dimension — Question 10

Section 9-8 · Physics pages 1310–1311

Question 10 A passive collision permits a bounded final velocity A 2 kg cart moving at 4 m/s overtakes a 1 kg cart moving at 1 m/s. The collision is isolated, has no internal energy release, and ends with separation or equal velocities. v2 (m/s) 3 5 e = 0 e = 1 Passive, separating outcomes span sticking through elastic separation. (a) Find the allowed interval of the lighter cart's final velocity. (b) Find the corresponding interval of kinetic-energy loss and identify both limiting cases.
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Question 10 - Solution A passive collision permits a bounded final velocity v2 (m/s) 3 5 e = 0 e = 1 (a) Derivation and calculation Let w = v2 −v1 ≥0. Conservation gives 2v1 + v2 = 2(4) + 1(1) = 9 ⇒v1 = 3 −w/3, v2 = 3 + 2w/3. A passive collision requires w ≤u1 −u2 = 3, hence 3 ≤v2 ≤5 m/s . (b) Result and interpretation With µ = 2/3 kg, L = 1 2µ(32 −w2) = 9 −w2 3 J ∈[0, 3] J . The endpoints are sticking (3, 3) with loss 3 J and elastic separation (2, 5) with zero loss. Physical check The center velocity remains 3 m/s throughout the allowed interval.

Question 10

A passive collision permits a bounded final velocity

A 2kg2\,\mathrm{kg} cart moving at 4m/s4\,\mathrm{m/s} overtakes a 1kg1\,\mathrm{kg} cart moving at 1m/s1\,\mathrm{m/s}. The collision is isolated, has no internal energy release, and ends with separation or equal velocities.

Diagram shown in the original SVG worksheet.

  1. Find the allowed interval of the lighter cart's final velocity.

  2. Find the corresponding interval of kinetic-energy loss and identify both limiting cases.

Question 10 - Solution

A passive collision permits a bounded final velocity

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Let w=v2−v1≥0w=v_2-v_1\geq0. Conservation gives 2v1+v2=2(4)+1(1)=9⇒v1=3−w/3,v2=3+2w/3.2v_1+v_2=2(4)+1(1)=9\Rightarrow v_1=3-w/3,\quad v_2=3+2w/3. A passive collision requires w≤u1−u2=3w\leq u_1-u_2=3, hence 3≤v2≤5m/s.\boxed{3\leq v_2\leq5\ \mathrm{m/s}}.

(b) Result and interpretation

With μ=2/3\mu=2/3 kg, L=12μ(32−w2)=9−w23J∈[0,3]J.\boxed{L=\tfrac12\mu(3^2-w^2)=\frac{9-w^2}{3}\ \mathrm J\in[0,3]\ \mathrm J}. The endpoints are sticking (3,3)(3,3) with loss 3 J and elastic separation (2,5)(2,5) with zero loss.

Physical check

The center velocity remains 3 m/s throughout the allowed interval.