# Chapter 5 - Worked Solutions

### to selected Complex Reasoning questions from the OUP text Senior Physics - Knowledge, Processes and Reasoning by Walding, Rapkins and Rossiter

Question 42
t = 0.19s uV = 0
sV = ut + 1/2 at2 = 0 + 1/2 x -10 x 0.192 = 0.18 m (18 cm)

Question 43
uH = 35 cos 20° = 32.9 ms-1
uV = 35 sin 20° = + 12.0 ms-1
sV = - 50 m
(a) vV2 = uV2 + 2as = (+12)2 + 2 x -10 x -50
vV = ± 33.8 ms-1 (must be negative, hence vV = - 33.8 ms-1)
vV = uV + at
- 33.8 = +12 + -10 t
t = 4.58 s

Using trigonometry: v (the hypotenuse) = 47.2 ms-1
The angle q = tan-1 (33.8/32.9) = 45.8° (to horizontal).

(c) sH = vH t = 32.9 x 4.58 = 150.7 m

Question 44
Maximum range: R = u2/a = 9.228 m
He was 0.328 m off maximum (96.5%).

Question 45
The diagram showing the 53° angle is as follows:

The trigonometric relationship between the velocities is as follows:

uV = v cos q ; uH = v sin q

(a) sV = uVt + 1/2 at2
-730 = uV x 4.5 + 1/2 x -10 x 4.52
-730 = 4.5 uV + -101
uV = -140 ms-1
uV also equal v cos 53° so: -140 = v cos 53°
v = -140/0.60 = -233 ms-1 at an angle of 53° downward

(b) sH = vH t = v sin 53° x 4.5
= 233 x 0.8 x 4.5 + 186.4 x 4.5 = 837 m

(c) vV = uV + at = -140 + -10 x 4.5 = - 185 ms-1
Using trigonometry:

v = 263 ms-1 (hypotenuse); q = tan-1 (185/186.4) = 44.8° to the horizontal

Question 46
(a) Ff = m FN so for a person to remain there, the friction must equal their weight (FW):
Ff = m FN = FW = mg

The normal force FN is equivalent to the centripetal force FC which equals mv2/r.
Hence:
m FN = mg
m FC = mg
m mv2/r = mg
m v2/r = g
v2 = gr/m = 10 x 2.1/0.4 = 52.5
v = 7.2 ms-1

(b)
FC = mv2/r = 49 x (7.2)2 / 2.1 = 1225 N

Question 47
(a) a = v2/r = 1202/800 = 18 ms-2 (=1.8 'g').
(b) 1.8 'g' < 4.5 'g' so he will remain conscious.
(c) T = FC + FW = mac + mg = 80 x 18 + 80 x 10 = 2240 N

Question 48
(a)
(i)

(ii)

(iii)

(iv)

(b)
(i) Same as (i) above except that the displacement gets less with each cycle.

(ii) Same as (ii) above except that the velocity gets less with each cycle.

(iii) Same as (iii) above except that the acceleration gets less with each cycle.

(iv) Spiral in towards the centre.

(c) The graph seems to be attracted to something at the origin of the two axes.

Question 49

Question 50

(a) sV = uVt + 1/2 at2
-20 = 0 + 1/2 x -10 t2
t = 2 s
sH = vH t = 145000/3600 x 2 = 80 m

(b) q = tan -1 80/20 = 76°

(c) vV = uV + at = 0 + -10 x 2 = -20 ms-1
vH = 40.3 ms-1

Impact
vector diagram: <img src = "ch5q50b.gif">
By Pythagoras' Theorem: v = 44.7° ; q = tan-1 20/40 = 27°

Launch
velocity: v = 80% x 44.7 = 35.8 ms-1; q = 27°
uV = v sinq = 35.8 sin 27° = 16 ms-1

Maximum height
is when vV = 0:
vV2 = uV2 + 2as
0 = 162 + 2 x -10 x s
sV = 12.8 m (maximum height at next bounce)

Range
= u2/a x sin 2q = 35.82/10 x sin (2 x 26.6° ) = 102 m

Alternatively by calculating time of flight:
vV = uV + at
-16 = +16 + -10 t
t = 3.2 s (time of flight)
sH = vH t = v cos 26.6° x 3.2 = 35.8 x cos 26.6° x 3.2 = 102 m