Chapter 5 - Worked Solutions

to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Question 6

The net is 0.9m above the launch height.
Horizontal:
t = shv = 4.3/vcos
q = 4.3/vcos55 = 7.5/v

Vertical:
sv = v sin55
t + at2
0.9 = 0.82 vt + -5t2
0.9 = 0.82 v (7.5/v) + -5 (7.5/v)2
0.9 = 6.15 + -5 (56.25/v2)
0.9 = 6.15 + -281.25/v2
5.25 = 281.25/v2
v2 = 281.25/5.25 = 53.57
v = 7.3 m s-1

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

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