Solving
Problems
[1] The position of a particle moving under uniform acceleration is some function
of time and the acceleration. Suppose we write this position s = kam tn where , k
is a dimensionless constant. Show by dimensional analysis that this expression
is satisfied if if m = 1 and n = 2
Solution
[s] [ka m t n ]
L Lm T 2m T n Lm T n 2m
m 1 and n 2m 0 n 2
[2] Newton’s law of universal gravitation is represented by
Mm
FG 2
r
Here F is the magnitude of the gravitational force exerted by one small object
on another , M and m are the masses of the objects, and r is a distance.
Force has the SI units kg ·m/ s2. What are the SI units of the proportionality
constant G?
Solution:
Mm Fr 2
FG 2 G
r Mm
kg.m / s 2 m 2 m3 N.m 2
G
kg kg kg.s 2
kg 2
[3] A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3.
From these data, calculate the density of lead in SI units (kg/ m3).
Solution
m 23 .94 10 3
11 .4 10 3 kg / m 3
V 2.10 10 6
One centimeter (cm) equals
0.01 m.
One kilometer (km) equals 1000
m.
One inch equals 2.54 cm
One foot equals 30 cm…
Example:
2.54cm in
(4.75in) (4.75in) 1 (4.75in) (4.75 2.54)cm 12.065cm
1in in
[4] If the rectangular coordinates of a point are given by (2, y) and its polar
coordinates are ( r , 30°), determine y and r.
Solution:
x 4 4 3
x r cos cos , then r
r 3 3
y 2 3
y r sin sin then y
r 3
[4] Two points in the xy plane have Cartesian coordinates (2.00, -4.00) m
and ( -3.00, 3.00) m. Determine (a) the distance between these points and
(b) their polar coordinates.
a) d (x 2 x1 ) 2 ( y 2 y1 ) 2
Solution:
d (3 2) 2 (3 4) 2 74m
b For (2,-4) the polar coordinate is (2,2√5) since
r x 2 y 2 4 16 2 5m
y
tan tan 1 2 269.5
x
[5] Vector A has a magnitude of 8.00 units and makes an angle of 45.0 ° with
the positive x axis. Vector B also has a magnitude of 8.00 units and is
directed along the negative x axis. Using graphical methods, find (a) the
vector sum A + B and (b) the vector difference A - B.
Solution:
[6] Given the vectors A = 2.00 i +6.00 j and B = 3.00 i - 2.00 j, (a) draw the
vector sum , C = A + B and the vector difference D = A - B. (b) Calculate C
and D, first in terms of unit vectors and then in terms of polar coordinates,
with angles measured with respect to the , +x axis.
Solution:
C A B 5i 4 j
4
tan 1 38.7
5
D A (B) i 8 j
tan 1 8 97 .2
[7] Consider the two vectors A = 3 i - 2 j and B = i - 4 j. Calculate (a) A + B,
(b) A - B, (c) │A + B│, (d) │A - B│, and (e) the directions of A + B and A - B.
Solution:
A B (3 1)i (2 4) j 2i 6 j
A B A (B) (3 1)i (2 4) j 4i 2 j
A B 4 36 40 2 10
A B 16 4 20 2 5
Direction of A+B tan 1 3 288
11
Direction of A-B
tan 26.6
2
[8] The vector A has x, y, and , z components of 8.00, 12.0, and -4.00 units,
respectively. (a) Write a vector expression for A in unit vector notation. (b)
Obtain a unit vector expression for a vector B four time the length of A pointing
in the same direction as A. (c) Obtain a unit vector expression for a vector C
three times the length of A pointing in the direction opposite the direction of A.
Solution:
a) A 8i 12 j 4k
b) B 4A 32i 48 j 16k
c) C 3A 24i 36 j 12k
[9] Three displacement vectors of a croquet ball are shown in Figure, where
A = 20.0 units, B = 40.0 units, and C = 30.0 units. Find the magnitude and
direction of the resultant displacement
Solution:
d x 40 cos45 30 cos45 49.5unit
d y 20 40 sin 45 30 sin 45 27unit
R (d x ) 2 (d y ) 2 56.4unit
d
tan 0.55 28.8
y
d x
[10] Find the magnitude and the direction of resultant force
Solution:
F x 25 15 cos53 8 cos240 30N
F y 10 15sin 53 8 sin 240 15N
R (Fx ) 2 (Fy ) 2 33.5N
tan
F y
0.5
F x
26.5