Ax,=: ae' cos (a)! + ), 0.8 = e'W => y= 4.5 x 103 s".
u: (E/p)" = (7.1 x 10/2700)"' = 5100 ms".
At fundamental Am = 41 = 4 m.
f=5|00/4 = 1.3 x 103112.
a) = 27;f= 8.1 x lOSrads'.
v =f/1wd, 5,1,. / A, = (1517 f => (57/ 1.
51 = I. (ch/f).
a = 1x
C Radioactivity and age of- the Eadh
cfw_a N - h'ne" M. = miginal numb
a = M.cfw_ l 6"
Therefore :1 - N e'( I e') 7.51:" I)
Son-v Ncfw_2- I) when ris half-life
mnzilsNaI-m "1] or Juan=nliNfeO IE-Illl -1) where lime r is in 109.
it) 2mn=mN[2"?'" -l')
E E.M. Induction
Method 1 Equatlng energy
Horizontal compancnt of magnetic eld B inducing em! in ring:
a = 44.5 x 10' cos I54
Magnum ux through ring at angle 3 = BMsin 9
where a = radius of ring
dip . t! sin wt .
Instantaneous emf: a Bxa- where a: = angul
Question Two ~ Solution
Focusing occurs for one "cyclotron" orbit of the electron.
Angular velocity w= e B / in; so time for one orbit T= 2 Irm / e B
Speed of electron u = (2 e V/ m)
Distance travelled D = T u cos [3: I u = (2],2 1t / B) (V m
B Heat Engine Question
T. in calculating work
(starts at TA) obtainabic,
we assume no loss
(friction etc.) in engine
AQ. = energy om body A
= -I71SAT| (AT, ve)
AQ; = "LSAT: (AT; +ve)
(slams at T3)
(a) For maximum amount of mechanical energy a
A Bungee Jumper
(a) The jumper comes to rest when
lost gravitational potential energy = stored strain energy
mgy = t k (y-I.)2
Icy: - 2y(kL + mg) + k1,2 = 0
This is solved as a quadratic.
y _ 2(kL +mg)i l4(kL + mg)Z - 41ch2
Need positive roo
(d) Assume T'> 4.50 x 10 and ignore I in both brackets:
00120 cfw_27240710 = cfw_277450 or 00120 cfw_$397621: cfw_6015.107
0020 = cfw_27 11- 1/0 710 "_. 2/(0122-1403"! = 2-1 13627
T 2 5.38 x 109 years
or 00120 633" T 1n0.01 = 114228
The jumper enters the SHM with the fall velocity = glr= 1[ZgL = U,
Period ofSHM = 27! k- = T
We represem a full SHM cycle by
Thejumper enters the SHM a time t given by
1 _, U 1 IZgL
r= sm -= sin
a) U (l) U
Jumper comes to rest at one halfcycle ofthe
Alter emerging from region between plates, electrons experience force due to
magnetic eld only. We approximate this by a vertical force. because angle of
electron to horizontal remains small.
Acceleration caused by this force a = B e u sin l ym
A shell with charge 4752550 moves om co to the surface of a sphere radius x
where the electric potential is
3 p Mo
and will therefore gain electrical potential energy ( p )(4m2p)6x
x=R 7 4 Z
_ 4 ,
D Spheriral charge
(a) Charge dcnsily = p = within sphere
x s R Field at disleneex:
421's"): " 4K,R'l
1: >3 Field at distanocx 'om the centre: E = Q 2
(b) Method 1'
Energy density is 805 2.
Energy in :1 mm shell ofthick
Method 2 Back Torque
Horizontal component ofmagnetic eld = B = 44.5 x 10' cos 64"
Cross-section of area ofring is A
Radius of ring = a
Density ofring = d
Resistivity = p
a) = angular velocity (a: positive when clockwise)
Resistance R = ~