Kinetic Energy — Question 9

Section 7-2 · Physics pages 888–889

Question 9 An energy budget constrains simultaneous speeds Use classical translational kinetic energy K = 1 2m|⃗v|2. Two independent carts A and B have masses 1 kg and 2 kg. Their total kinetic energy is exactly 36 J in the same frame. Each must have speed at least 2 m/s. A: 1 kg vA ≥2 m/s B: 2 kg vB ≥2 m/s Both carts share a fixed kinetic-energy budget of 36 J. (a) Find the greatest possible speed of A. (b) Find the greatest possible speed of B. (c) Can both maxima occur simultaneously? Give the feasible relation between their speeds.
Show solutionHide solution
Question 9 - Solution An energy budget constrains simultaneous speeds A: 1 kg vA ≥2 m/s B: 2 kg vB ≥2 m/s (a) Reserve B’s minimum energy. The governing relation is 1 2mAv2 A + 1 2mBv2 B = 36 J B requires at least 1 2(2)22 = 4 J. At most 32 J remains for A, so vA,max = √ 2(32)/1 = 8 m/s, attained when vB = 2 m/s. (b) Reserve A’s minimum energy. A requires 2 J; B can then have 34 J. Thus vB,max = √ 34 m/s, attained when vA = 2 m/s. (c) Impose one joint constraint. The feasible speeds obey 1 2v2 A + v2 B = 36, with both at least 2. The two separate maxima would require 32 + 34 = 66 J, so they cannot be attained together.

Question 9

An energy budget constrains simultaneous speeds

Use classical translational kinetic energy K=12m|v→|2K=\tfrac12 m|\vec v|^2. Two independent carts A and B have masses 1 kg and 2 kg. Their total kinetic energy is exactly 36 J in the same frame. Each must have speed at least 2 m/s.

Diagram shown in the original SVG worksheet.

  1. Find the greatest possible speed of A.

  2. Find the greatest possible speed of B.

  3. Can both maxima occur simultaneously? Give the feasible relation between their speeds.

Question 9 - Solution

An energy budget constrains simultaneous speeds

Diagram shown in the original SVG worksheet.

(a) Reserve B’s minimum energy.

The governing relation is 12mAvA2+12mBvB2=36J\boxed{\tfrac12m_Av_A^2+\tfrac12m_Bv_B^2=36\ \mathrm J} B requires at least 12(2)22=4J.\tfrac12(2)2^2=4\ \mathrm J. At most 32 J remains for A, so vA,max=2(32)/1=8m/s,v_{A,\max}=\sqrt{2(32)/1}=8\ \mathrm{m/s}, attained when vB=2m/s.v_B=2\ \mathrm{m/s}.

(b) Reserve A’s minimum energy.

A requires 2 J; B can then have 34 J. Thus vB,max=34m/s,v_{B,\max}=\sqrt{34}\ \mathrm{m/s}, attained when vA=2m/s.v_A=2\ \mathrm{m/s}.

(c) Impose one joint constraint.

The feasible speeds obey 12vA2+vB2=36,\tfrac12v_A^2+v_B^2=36, with both at least 2. The two separate maxima would require 32+34=66J,32+34=66\ \mathrm J, so they cannot be attained together.