By A. Mous

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3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. (y )2 = (1 − y 2 )(1 − k 2 y 2) y + y(1 + k 2 ) − 2k 2 y 3 = 0 y = snu, y = (cnu)(dnu) (y )2 = (1 − y 2 )(1 − k 2 + ky 2) y + y(1 − 2k 2 ) + 2k 2 y 3 = 0 y = cnu, y = (−snu)(dnu) (y )2 = (1 − y 2 )(y 2 − 1 + k 2 ) y + y(k 2 − 2) + 2y 3 = 0 y = dnu, y = −k 2 (snu)(cnu) (y )2 = (1 + y 2 )(1 + (1 − k 2 )y 2) y − y(2 − k 2 ) − 2y 3(1 − k 2 ) = 0 y = tnu, y = (dnu)/(cn2 u) (y )2 = (y 2 − 1)(y 2 − k 2 ) y + y(1 + k 2 ) − 2y 3 = 0 y = 1/snu, y = −(cnu)(dnu)/(sn2 u) (y )2 = (y 2 − 1)[(1 − k 2 )y 2 + k 2 ] y + y(1 − 2k 2 ) − 2y 3(1 − k 2 ) = 0 y = 1/cnu, y = (snu)(dnu)/(cn2 u) y = (1 + y 2 )2 − 4k 2 y 2 y + 2y(2k 2 − 1) − 2y 3 = 0 y = (dnu)(tnu) (y )2 = 4y(1 − y)(1 − k 2 y) y − 2 + 4y(k 2 + 1) − 6k 2 y 2 = 0 y = sn2 u (y )2 = 4y(1 − y)(1 − k 2 + k 2 y) y − 2(1 − k 2 ) − 4y(2k 2 − 1) + 6k 2 y 2 = 0 y = cn2 u (y )2 = 4y(1 − y)(y − 1 + k 2 ) y − 2(k 2 − 1) − 4y(2 − k 2 ) + 6y 2 = 0 y = dn2 u (y )2 = 4y(y − 1)(y − k 2 ) y − 2k 2 + 4y(k 2 + 1) − 6y 2 = 0 y = 1/(sn2 u) (y )2 = 4y(y − 1)[(1 − k 2 )y + k 2 ] y + 2k 2 − 4y(2k 2 − 1) − 6y 2(1 − k 2 ) = 0 y = 1/(cn2 u) (y )2 = 4y[(1 + y)2 − 4k 2 y] 53 14.

Show that the time required for the object to reach the origin at the sun is t = π( mre3 1/2 ) . 8k What did dimensional analysis tell you to expect (see Appendix B)? If the object is the earth, how long does it take? 59 3. A particle of mass m is initially at rest. A constant force F0 acts on the particle for a time t0 . The force then increases linearly with time such that after an additional interval t0 the force is equal to 2F0 . Show that the total distance the particle goes in the total time 2t0 is, 13 2 F0 t0 /m.

4. A boy runs and then slides on some slushy ice. It is hypothesized that friction varies as the square root of the speed, F (v) = −cv 1/2 . The initial speed of the boy is v0 at time t = 0. What does dimensional analysis give for a length scale for the problem? Find the values of v and x as function of the time t. Partial answer: 1/2 c2 3 cv0 2 )t + t. x = v0 t − ( 2m 12m2 Show that the boy cannot travel farther than xmax = 2m 3/2 v . 3c 0 Use some reasonable values of m, v0 and xmax to estimate c in SI units.

### A Short Introduction to Theoretical Mechanics by A. Mous

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