Chapter 5 - Worked Solutions

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

The net is 0.9m above the launch height.
Horizontal:
t = s_hv = 4.3/vcosq = 4.3/vcos55° = 7.5/v

Vertical:
s_v = v sin55°t + ½ 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

t = 0.19s u_V = 0
s_V = ut + 1/2 at² = 0 + 1/2 x ^-10 x 0.19² = 0.18 m (18 cm)

u_H = 35 cos 20° = 32.9 ms^-1
u_V = 35 sin 20° = ⁺ 12.0 ms^-1
s_V = - 50 m
(a) v_V² = u_V² + 2as = (⁺12)² + 2 x ^-10 x ^-50
v_V = ± 33.8 ms^-1 (must be negative, hence v_V = - 33.8 ms^-1)
v_V = u_V + 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) s_H = v_H t = 32.9 x 4.58 = 150.7 m
 

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

The diagram showing the 53° angle is as follows:


The trigonometric relationship between the velocities is as follows:


u_V = v cos q ; u_H = v sin q
 
(a) s_V = u_Vt + 1/2 at²
-730 = u_V x 4.5 + 1/2 x ^-10 x 4.5²
-730 = 4.5 u_V + ^-101
u_V = ^-140 ms^-1
u_V also equal v cos 53° so: ^-140 = v cos 53°
v = -140/0.60 = ^-233 ms^-1 at an angle of 53° downward
 
(b) s_H = v_H t = v sin 53° x 4.5
= 233 x 0.8 x 4.5 + 186.4 x 4.5 = 837 m
 
(c) v_V = u_V + 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
 

(a) F_f = m F_N so for a person to remain there, the friction must equal their weight (F_W):
F_f = m F_N = F_W = mg
 
The normal force F_N is equivalent to the centripetal force F_C which equals mv²/r.
Hence:
m F_N = mg
m F_C = 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)
F_C = mv²/r = 49 x (7.2)² / 2.1 = 1225 N
 

(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 = F_C + F_W = ma_c + mg = 80 x 18 + 80 x 10 = 2240 N
 

(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.
 

(a) s_V = u_Vt + 1/2 at²
-20 = 0 + 1/2 x ^-10 t²
t = 2 s
s_H = v_H t = 145000/3600 x 2 = 80 m
 
(b) q = tan ^-1 80/20 = 76°
 
(c) v_V = u_V + at = 0 + ^-10 x 2 = ^-20 ms^-1
v_H = 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°
u_V = v sinq = 35.8 sin 27° = 16 ms^-1
 
Maximum height is when v_V = 0:
v_V² = u_V² + 2as
0 = 16² + 2 x ^-10 x s
s_V = 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:
v_V = u_V + at
-16 = +16 + ^-10 t
t = 3.2 s (time of flight)
s_H = v_H t = v cos 26.6° x 3.2 = 35.8 x cos 26.6° x 3.2 = 102 m

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