Problem#1
The position vector for an electron isr = (5.0 m)i – (3.0 m)j + (2.0 m)k.
(a) Find the magnitude of r, and
(b) Sketch the vector on a right-handed coordinate system.
Answer:
(a) The magnitude of r is
r = √(x2 + y2 + z2)
r = √[(5.0 m)2 + (–3.0 m)2 + (2.0 m)2] = 6.2 m
(b) Sketch the vector on a right-handed coordinate system
Problem#2
A watermelon seed has the following coordinates: x = 5.0 m, y = 8.0 m, and z = 0 m. Find its position vector
(a) in unit-vector notation and as
(b) a magnitude and
(c) an angle relative to the positive direction of the x axis.
(d) Sketch the vector on a right-handed coordinate system.
If the seed is moved to the xyz coordinates (3.00 m,0 m, 0 m), what is its displacement,
(e) in unit-vector notation and as
(f) a magnitude and (g) an angle relative to the positive x direction?
Answer:
(a) The position vector, according to r = xi + yj is
r = (–5 m)i + (8.0 m)j
(b) The magnitude is
r = √(x2 + y2)
r = √[(–5.0 m)2 + (8 m)2] = 9.4 m
(c) that the vector lies in the xy plane and we obtain
θ = tan-1(y/x) = tan-1(8m/-5.0 m) = –580 or 1220
where the latter possibility (122° measured counterclockwise from the +x direction) is chosen since the signs of the components imply the vector is in the second quadrant.
(d) The sketch is shown to the right. The vector is 122° counterclockwise from the +x direction.
(e) The displacement is
∆r = r’ – r
where r is given in part (a) and r’ = (3.0 m)i. Therefore,
∆r = (8.0 m)i + (–8.0)j
(f) The magnitude of the displacement is
|∆r| = √[(8.0 m)2 + (–8.0 m)2] = 11 m.
(g) The angle for the displacement,
tan-1 (y/x) = tan-1 (-8 m/8 m) = –45.00 or 1350.
where we choose the former possibility (−45°, or 45° measured clockwise from +x) since the signs of the components imply the vector is in the fourth quadrant. A sketch of Δr is shown on the right
Problem#3
A positron undergoes a displacement ∆r = 2.0i – 3.0j + 6.0k, ending with the position vector r = 3.0i - 4.0k, in meters. What was the positron’s initial position vector?
Answer:
The initial position vector ro satisfies ∆r = r – r0, which results in
r0 = r – ∆r
r0 = (3.0i - 4.0k) m – (2.0i – 3.0j + 6.0k) m = (–2.0i + 6.0j – 10.0k) m
Problem#4
The minute hand of a wall clock measures 10 cm from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the tip are to be determined for three time intervals. What are the
(a) magnitude and
(b) angle from a quarter after the hour to half past, the
(c) magnitude and
(d) angle for the next half hour, and the
(e) magnitude and
(f) angle for the hour after that?
Answer:
We choose a coordinate system with origin at the clock center and +x rightward (toward the “3:00” position) and +y upward (toward “12:00”).
(a) In unit-vector notation, we have r1 = (10 cm)i and r2 = (–10 cm)j. then
∆r = r – r0
∆r = (–10 cm)j – (10 cm)i
The magnitude is given by
|∆r| = √[(–10 m)2 + (–10 m)2] = 14 m.
(b) the angle is
tan-1 (y/x) = tan-1 (-10 m/-10 m) = 45.00 or –1350.
We choose −1350 since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write
∆r = r2 – r1 = (−10 cm)i + (−10 cm) j → (14 cm ∠ − 1350).
(c) In this case, we have r1 = (−10 cm) j and r2 = (10 cm)i, and Δr = (20 cm)j. Thus, |∆r| = 20 cm.
(d) the angle is given by
θ = tan-1(20 cm/0) = 900.
(e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero.
(f) The corresponding angle for a full-hour sweep is also zero.
Post a Comment for "Position and Displacement Problems and Solutions"