Answers:
Multiple Choice
1. B
2. C
3. C
4. A
5. B
6. E
7. G
8. E
9. A
10. B
Worked Problems
11. A) 3.0 m B) 1.2 (or 1.3) m/s C) 2.0 m/s D) 2.7 s
12. A) 217 m/s B) 52.9 deg C) 1530 m
13. (2.15 i + 2.66 j) m/s2 = 3.42 m/s2 at 51.1 deg
14. A) 4.14 m/s2 B) up the incline (or equivalent)
15. A) 27.0 N B) 0.555
16. 0.940 N
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Explanations:
1. You only have information on the beginning and ending positions,
and not on the path taken, hence you cannot know the total
distance traveled and hence cannot know the speed. The average
acceleration requires that you know the change in velocity, but
no information on velocity is given. You can, however, compute
the average velocity = (rf - ri)/(tf
- ti).
(see Chap 4, Q 3).
2. The acceleration due to gravity (near Earth) is always
9.8 m/s2 down. With the coordinates specified, down
is (-j). Hence the correct answer is C.
(see Chap 4, Q 17)
3. The average acceleration is the change in velocity divided
by the change in time. The initial velocity is vt
to the right, the final velocity is vt to the left.
The magnitude of the change in velocity is 2 vt.
(see Chap 4, definition of average acceleration).
4. The x-component of the velocity of the object is given
by dx/dt and is correctly shown in answer C. Since the
x-component is not zero at t = 0, the velocity of the object
cannot be zero at t = 0, no matter what the y-component may be.
Hence the answer is A.
(see Chaps 2 and 3)
5. At the peak, the y-component of the velocity is zero. For
projectile motion the x-component of the velocity is constant.
Hence, at the peak the velocity is vx i.
(see Chap 4, Q 17).
6. One eliminates all the answers except E, which is "none of
the above." For A, the net force must be zero, but that doesn't
mean there aren't any forces. In fact you know that at least
gravity is acting on the book. For B, the normal force is
perpendicular to the table's surface and will not be vertical.
For C, the coefficient of friction is never negative. For D,
the book is "at rest" and hence the acceleration is zero.
(see Chap 5, e.g. Q 10 and fig. 23)
7. 1 L = (10 cm)3 = (0.1 m)3 = 10-3
m3
So 1 mL = 10-3L = 10-6 m3.
Hence 1 m3 is 106 mL.
(see Chap 1)
8. m2 has the following forces: gravity (down), normal force from surface (up), contact force from m1 (to right), and friction (to left). Four forces. (See chap 5, for example, figure 23)
9. Since m1 is "at rest", the net force on it is
zero. Hence the tension in the rope is equal in magnitude to
gravity pulling down on m1, T = m1 g . Since we have
massless, frictionless pulleys, the tension in the rope is
the same everywhere. Hence the tension on the ceiling is
m1 g .
(see Chap 5)
10. Between 1s and 3s the paths are straight lines. Since
v = dx/dt is a constant for a straight line, the acceleration
is zero there. Between 0s and 1s, the path on the x vs t
diagram is curved (upward), indicating an acceleration is
present (the velocity is changing). The only answer which fits
this is B.
(see Chap 2, P 3 and example 2.3 on page 32)
11. A) Displacement is the change in position. At 3s the
object is at about 3.5 m and at 1s it is at about 0.5m. Hence
the change is 3.0 m.
B) The average velocity is the displacement divided by how
long it took. Between 0 and 1s, the object moves from about
-0.75 to +0.5m. The change is then +1.25 m during this 1s.
Hence the average velocity is 1.25 m/s (to 2 sig figs, this
would be 1.3 m/s or 1.2 m/s depending on the exact values
you read off the graph).
C) The velocity is the slope of the line. Near 1.5 s the
line is straight and x is changing by +1 m for each 0.5 s.
Hence the velocity is 2.0 m/s.
D) The velocity is zero when the curve is horizontal. That
occurs here only between 2.5 and 3.0 s, at about 2.7s.
(See Chap 2)
12. A) The magnitude is the square root of the sum of the
squares of the components or the square root of
( (131)2 + (173)2 ) m/s .
B) The angle up from horizontal (the x-direction) is the
arctangent of the y-component of the
velocity divided by the x-component of the velocity, or
theta = tan-1(173/131).
C) This can be computed a number of ways. The fastest is
to use vyf2 = vyi2
- 2 g h . Here vyf = 0 and
vyi = 173 m/s. Now solve for h. Alternatively,
you can find the time to get to the peak using vyf =
vyi - gt , then use the time, t, to find h.
(Chap 4, projectile motion, Chap 3 for mag. and dir. of vector)
13. In component form:
F1 = 12 N i
F2 = 10 N (cos(120 deg) i + sin(120 deg) j)
where 120 deg is the angle counter-clockwise from the x-axis.
Add these two together (by components) and divide by m to get the acceleration.
(See Chap 5, P 15)
14. A) The component of gravity along the incline is mg sin(25 deg)
with a direction down the incline. Hence the magnitude of the
acceleration is g sin(25 deg).
B) When the block is going down the incline, friction resists
this motion and hence its direction is up the incline. (any
answer which is equivalent to "up the incline" is ok).
(See Chap 5, P32)
15. A) Set up F = ma for each mass. For m2 you have
m2g down and T = tension, up. For m1
you have T to the right and friction = 0.243 m1g to
the left. When m2 goes down, m1 goes to
the right. So you have three equations (two F = ma, and the
"connection" between the motion of the two masses. Solve these
for the tension.
B) The largest the force of friction can be is musm1g.
where mus is the coefficient of static friction. If
m2 is not moving, the tension = m2g, and
hence one needs
m2g = musm1g
now solve for mus.
(I have spelled out "mu" here since greek letters are problematic in html files).
(See Chap 5, P49, P67)
16. The accelerations of m1 and m2 are equal.
Hence, a = F/(m1 + m2). For m2,
have Fnet = m2 a . Substitute in the acceleration and
you get the net force on m2.
(See Chap 5, P 56)