Shaskolskaya & Eltsin Problem 96

Physics · mechanics · momentum & collisions

Problem

The principle of relativity demands that the same physical laws should obtain in the two systems under consideration, and in particular the law of conservation of energy, according to which the change of energy of a body must equal the work done by external forces. Therefore in system I the following relationship must obtain m2(v22−v12)=Fs\frac{m}{2}(v_2^2 - v_1^2) = Fs, (1) where s is the distance travelled by a body in system I in the time during which its velocity rises from v1v_1 to v2v_2. Correspondingly in system II m2(v22−v12)−mv(v2−v1)=Fs1\frac{m}{2}(v_2^2 - v_1^2) - mv(v_2 - v_1) = Fs_1, (2) where s1s_1 is the distance travelled by a body in system II in the same time. But since the velocity of a body in systems I and II is not the same, neither are s and s1s_1 the same. In fact, since a body moves under the impulse of a force F, with acceleration F/m, in system I we have: s=v1t+Fm⋅t22s = v_1t + \frac{F}{m} \cdot \frac{t^2}{2} and in system II correspondingly s1=(v1−v)t+Fm⋅t22s_1 = (v_1 - v)t + \frac{F}{m} \cdot \frac{t^2}{2} Therefore s−s1=vts - s_1 = vt. But since F/m=a=(v2−v1)/tF/m = a = (v_2 - v_1)/t, t=v2−v1⋅mFt = \frac{v_2 - v_1 \cdot m}{F} and s−s1=v(v2−v1)⋅mFs - s_1 = \frac{v(v_2 - v_1) \cdot m}{F}, and therefore F(s−s1)=mv(v2−v1)F(s - s_1) = mv(v_2 - v_1). Thus the work of outside forces in system I is as much greater than the work of outside forces in system II as the change in kinetic energy in system I is greater than the change in kinetic energy in system II. And since the change in energy in system I equals the work done by external forces, this also holds for system II. Therefore Newton's principle of relativity is not broken. V. The Dynamics of Motion in a Circle

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