Answers
Answers to assigned even numbered problems appear below:
Ch.1. 1:2a) 9.14x10^1m, b) 2x10^3m, c) 4.2194x10^4m;
1:4a) 3.523min, b) 391.4sec; 1:12a) 25m/s, b) 88km/hr;
1:16a) 8.4x10^3sec, b) 1.40x10^2min, c) 2.33hr;
Ch. 2. 2:6) 0->1 is 2.7m/s, 1->2 is 18.9m/s, 2->3
is 51.3m/s, 3->4 is 99.9m/s; 2:16a) B, b) A&B, c)
A, d) A, e) A.
Ch. 3. 3:4a) 0.38sec, b) 0.48sec;
3:6a) A&C, b) B, c) A, B &C, d) A, B&C, e) B;
3:8) C; 3:13) 7.5m/s^2; 3:14) 12m/s^2
3:18) D; 3:20) E;
Ch. 4. 4:2a) 10.4m/s, b) -9.2m/s;
4:10) accel. is neg. as soon as feet leave floor; 4:16a)
0.86sec, b) 0.43sec; 4:22) 30.6m;
Ch. 5. 5:2) a and c; 5:6) 60m;
5:20) 22.5%
Ch. 6. Only odd # problems assigned.
Ch. 7. 7:2) 22m/s; 7:4) runner
should be safe; 7:20) 4.9m/s; 7:24a)
same for all, b) same for all, c) C, d) C, e) A.
Ch. 8. 8:2) weight does not cease, but contact force
does; 8:4a) N>W, b) N=0, c) N>W; 8:6)
mg is exerted by earth; 8:8a, b, c and f) yes, d) not
necessarily, e) no; 8:14a) 588N; 8:16a)
different diections, b) directions must be opposite.
Ch. 9. 9:6) N and mg are not action/reaction pair;
9:14)T(W)=0, T(P)=56Nm, T(N)=112Nm, T(f)=56Nm; 9:22)
eqlb. not possible because T exerts both net force and torque on skier.
Ch. 10. 10:2) CM can go below the bar by bending the bar around
the body as one goes over. In this case the CM lies outside the body
itself. 10:8) Since the body's density is different from that
of water the center of bouyancy is different that the center of gravity.
10:18) When the swimmer is upside down in the water, the torques due to
B and C would tend to take one away from equilibrium. On the other
hand, with the swimmer right side up these same torques bring one back
to eqlb.
Ch. 11. 11:6) Greater mass and muscle movement in starting a
sprint translates to a greater reaction time.
Ch. 12. 12:24) 21m/s^2.
Ch. 13. 13:4a) 45 degrees, b) If Nx is less (greater) than mg,
then theta decreases (increases); 13:6) 14m/s.
Ch. 16. 16:2) Not treated as a collision because the contact
force does not dwarf the other forces on the ball and the time of contact
is not short; 16:4) 30N.s; 16:8) For
a given force the impulse delivered can be increased by increasing the
time over which that force acts; 16:10a) 25N.s,
b) 75N.s, c) 125N.s, d) 150N.s; 16:18) change
in momentum = 2*Vi, away from the wall.
Ch. 17. 17:8) deltaP(club) = - deltaP(ball) and J(club)
= - J(ball); 17:10) 7.4m/s; 17:14a)
-50m/s, b) 30m/s, c) 0.6; 17:16) 0.25m;
17:20a) 0.25, b) 0.5, c) 0.75.
Exam 1
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Exam 2
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Final Exam
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