Irodov – Problems in General Physics

Fig. 1.

(a)

A

3

2

1

g

-1

5

5

4

1 2 3 4

ANSWERS AND SOLUTIONS

1A. v = 112-c = 3.0 km per hour.

1.2. (v) = 2v 0 (v 1 + v 2 )/(2vo + v1 + v2).

1.3. At= -1/ 1 — 4 (v)/urt 15 s.

1.4. (a) 10 cm/s; (b) 25 cm/s; (c) to = 16 s.

1.5. (r 1 — r 2 )/1 r 1 r 2 1 == (v -- vi)/1 v2 (^2) -- vl
1.6. v' = V 4 + v2 + 2vov cos q)^ 40 km per hour, cp' = 19°.
1.7. u— (1—va/v'2
vo
)- 1 / - 2
1 =3.0 km per hour.
1.8. TA/TB =1/1/ 11 2— 1 -- 1.8.
1.9. A=aresin (1/n) + a/2 -- 120°.
1.10. / = vot y 2 (1— sin 0) = 22 m.
1.11. 1 = (vi+ v 2 )-liviv 2 1g = 2.5 m.
1.12. t = 2a/3v.
1.13. It is seen from Fig. 1a that the points A and B converge
with velocity v — u cos a, where the angle a varies with time. The
points merge provided the following two conditions are met:
S
(V U cos cc) dt=1, vcosa at = UT,
0 0
where .t is the sought time. It follows from these two equations that.
T =vl/(v 2 -
1.14. x 1 — x 2 = 1— /VT (t + T12) = 0.24 km. Toward the train
with velocity V = 4.0 m/s.
1.15. (a) 0.7 s; (b) 0.7 and 1.3 m respectively.