Download Test 1

Test 1
phy 240
1. a) What are primitive concepts? Give an example.
b) Write all four conditions which must be satisfied by a scalar product. (Use
different symbols to distinguish operations on vectors from operations on
numbers.)
c) Define the magnitude of a vector.
d) What does it mean that multiplication of a vector by a number is distributive
over addition of vectors? (Use different symbols to distinguish operations on
vectors from operations on numbers.)
2. The Cartesian coordinates of a point (scalar components of the position vector)
are r = (−3.5m, −2.5m).
a) Identify the scalar components of this vector.
b) Identify the vector components of this vector.
c) Find the polar coordinates of this point.
3. The one kilogram standard is a platinum-iridium cylinder 39 mm in height and
39 mm in diameter. What is the density of the material?
4. Two vectors A and B have precisely equal magnitudes. For the magnitude of
A+B to be 100 times greater than the magnitude of A-B what must the angle
between them be?
-1-
a) Primitive concepts are not formally defined. Understanding of these concepts
is based on intuition. Mass, length, time are examples of primitive concepts.
b) The following conditions are satisfied by scalar product:
1. A o B = B o A
2. (α • A ) o B = α(A o B)
3. (A ⊕ B ) o C = (A o C) + (B o C)
(commutative)
(mixed associative)
(distributive over addition of vectors)
4. A2 ≥ 0
A2 = 0 if and only if A = 0
c) The magnitude of a vector is defined as the square root of the scalar square of
the vector:
A = A2
If α is an arbitrary number, and A and B are two arbitrary vectors then
α • (A ⊕ B ) = α • A ⊕ α • B
-2y
rx
θ
j
i
x
ry
r
a) The scalar components of the position vector r are given in the problem:
x = −3.5 m
y = −2.5 m
b) The vector components are:
rx = (−3.5m) i
ry = (−2.5m) j
c) Following the definition, the magnitude of this vector is
r = r2 =
(− 3.5m)2 + (− 2.5m)2 = 4.3m
The Cartesian components are also related to the angle θ indicating the
direction of the vector
tan θ =
y − 2.5m
=
x − 3.5m
Since the vector is in the third quadrant, the angle with the x-axis is
θ = 180° + arctan
− 2.5m
= 216°
− 3.5m
-3-
39.0 mm
39.0 mm
From the definition of density ρ, for a uniform object, its mass m is directly
proportional to the volume V
1)
m=
∫ ρdV = ρ ⋅ ∫ dV = ρV
object
object
The volume of a cylinder depends on its radius r and height h
2)
V = πr 2 h
The rest is algebra:
ρ=
m
m
1kg
kg
=
=
≈ 2.15 × 104
2
V πr 2 h
m3
π ⋅ 19.5 × 10− 3 m ⋅ 39.0 × 10− 3 m
(
) (
)
-4-
A+B
A
A-B
B
θ
-B
Solution 1
Let θ represent the angle between the directions of the two vectors. Since A
and B have the same magnitudes, A, B, and A+B form an isosceles triangle in
which the angles are (π-θ), θ/2 and θ/2. Similarly A, B, and A-B form an isosceles
triangle in which the angles are θ, (π-θ)/2 and (π-θ)/2. From the law of cosines,
the magnitude of both A+B and A-B can be expressed in terms of the magnitude of
vector A (and B)
A + B = A 2 + B2 − 2AB cos(π − θ ) = 2A 2 (1 + cosθ ) = 2A cos
A − B = A 2 + B2 − 2AB cosθ = 2A2 (1 − cos θ) = 2A sin
θ
2
θ
2
The ratio given in the problem yields the following trigonometric equation
tan
θ
1
=
2 100
from which
θ = 2 ⋅ arctan
1
≈ 0.02 ≈ 1.15°
100
(Solution 2
One can avoid using the law of cosines. From the definition of magnitude
θ
A + B = (A + B )2 = A 2 + B 2 + 2A o B = 2A 2 (1 + cosθ) = 2A cos
2
θ
A − B = (A − B)2 = A2 + B2 − 2A o B = 2A2 (1 − cosθ) = 2A sin
)
2