英语翻译A A bungee jumper is attached to one end of a long elast

The jumper comes to rest when
lost gravitational potential energy = stored strain energy
mgy = k (y-L)2
0.1
ky2 – 2y(kL + mg) = 0
0.1
This is solved as a quadratic.



=
0.2
Need positive root; lower position of rest (other root after initial rise).
0.1
0.5
The maximum speed is attained when the acceleration is zero and forces balance; i.e. when mg = kx 0.1
Also kinetic energy = lost potential energy – strain energy within elastic rope
0.1
0.1
0.2
0.5
Time to come to rest = time in free fall + time in SHM of rope to stop stretching
0.1
Length of free fall =
Therefore
0.2
The jumper enters the SHM with free fall velocity = gtf ==
0.1
Period of SHM = = T
0.1
We represent a full SHM cycle by
The jumper enters the SHM at time  given by
 = =
0.2
Jumper comes to rest at one half cycle of the SHM at total time given by
= tf + (T/2 - 

=-
= -
={ - }
This is the same as
={ }
= tan-1{-}
0.2
1.0
B Heat Engine Question
In calculating work obtainable,
we assume no loss (friction etc.) in engine working.
Q1 = energy from body A
= -msT1 (T1 –ve)
Q2 = msT2 (T2 +ve)
(a) For maximum amount of mechanical energy assume Carnot engine
= throughout operation (second law)
0.2
ButQ1 = -msT1 and Q2 = msT2
0.2
=
0.1
ln= ln
0.1
0.2
0.8 Q1 = -ms= ms(TA – T0)
0.2
Q2 = ms= ms(T0 – TB)
0.1
W = Q1 – Q2
0.2
W = ms(TA-T0 –T0 + TB) = ms(TA+TB – 2T0) = ms (TA+TB - 2)
or ms
0.2
0.7

(d) Numerical example:
Mass = volume density
W = 2.50 1.00 103 4.19 103 (350 + 300 - 2) J
= 20 106 J
= 20 MJ
0.5
0.5
C Radioactivity and age of the Earth
N = N0e- N0 = original number
0.1
n = N0(1 – e-t)
0.1
Therefore n = N et(1 – e-t) = N(et – 1)
0.1
So n = N(2t/ – 1) where  is half-life
or as  = , n = N()
0.1
or where time t is in 109 years
0.1
0.5
(b) or
0.1
0.1
(c) In mixed uranium (i.e. containing Pb of both natural and radioactive origin)
204 : 206 : 207 have proportions 1.00 : 29.6 : 22.6
In pure lead (no radioactivity) 1.00 : 17.9 : 15.5
Therefore for radioactively produced lead by subtraction
206 : 207 11.7 : 7.1
0.2
Dividing equations from (a) and (b) gives
or
0.1
or
0.1
0.0120 {2T/0.710 – 1} = {2T/4.50 - 1}
or 0.0120 {e0.9762T – 1}= {e0.1540T –1}
0.1
0.5
(d) Assume T >> 4.50 109 and ignore 1 in both brackets:
0.2
0.0120 {2T/0.710 } = {2T/4.50} or 0.0120 {e0.9762T}= {e0.1540T}
0.0120 = {2T/4.50 - T/0.710 }= 2T(0.222-1.4084) = 2-1.1862T
T = = 5.38
T = 5.38 109 years

or 0.0120 = e-0.8222T T =
T = 5.38 109 years
0.2
0.4
T is not >> 4.50 109 years but is > 0.71 109 years
0.1
We can insert the approximate value for T (call it T* = 5.38 109 years) in the 2T/4.50 term and obtain a better value by iteration in the rapidly changing 2T/0.710 term). We now leave in the –1’s, although the –1 on the right-hand side has little effect and may be omitted).
0.1
Either 0.0120(
T = 0.710
0.2
Put T* = 4.80(0) 109 years
T = 0.710
Further iteration gives 4.52
0.1
or
0.0120(e0.9762T –1) = (e0.1540T* - 1) and similar
So more accurate answer for T to be in range 4.6 109 years to 4.5 109 years (either acceptable).

0.5D Spherical charge

(a) Charge density = within sphere
0.3
x R Field at distance x:
E ==
0.3
x R Field at distance x from the centre:
0.2
0.8
(b) Method 1
Energy density is .
0.1
x R
Energy in a thin shell of thickness x at radius x is given by
= = x 0.1
Energy within the spherical volume = =
0.2
x > R
Energy within spherical shell = = x
0.1
Energy within the spherical volume for x R
= =
0.2
Total energy associated with the charge distribution = +
= =
0.1
0.8

Method 2
A shell with charge 4x2x moves from to the surface of a sphere radius x
0.1
where the electric potential is
0.2
and will therefore gain electrical potential energy ()(4r2x
0.1
Total energy of complete sphere =
0.2
Putting Q = charge on sphere = ,
So that total energy is = =
0.2
0.8
(c) Binding energy Ebinding = Eelectric - Enuclear
0.1
Binding energy is a negative energy
Therefore -8.768 = Eelectric - 10.980 MeV per nucleon
Eelectric = 2.212 MeV per nucleon
0.1
Radius of cobalt nucleus is given by R =
= m
= 5.0 10-15 m
0.2
0.4
E E.M. Induction
Method 1 Equating energy
Horizontal component of magnetic field B inducing emf in ring:
B = 44.5 10-6 cos 64o 0.2
Magnetic flux through ring at angle  = Ba2sin 
where a = radius of ring 0.1
Instantaneous emf = = Ba2 where  = angular velocity
= Ba2cos t= Ba2cos  
R.m.s. emf over 1 revolution =  0.2
Average resistive heating of ring = 0.1
Moment of inertia = 0.1
Rotational energy = where m = mass of ring 0.1
Power producing change in  = {} =
0.1
Equating: = -
0.1
= -dt
0.1
If T is time for angular velocity to halve,
0.1
ln 2 = T 0.2
But R = where A is cross-sectional area of copper ring 0.1
m = 2ad A (d = density) 0.1
ln 2 = = 0.1

T = =
= 1.10(2) 106 s (=306 hr = 12 days 18 hr) 0.2
2.0(Part E)
Method 2 Back Torque
Horizontal component of magnetic field = B = 44.5 10-6 cos 64o 0.2
Cross-section of area of ring is A
Radius of ring = a
Density of ring = d
Resistivity = 
= angular velocity positive when clockwise)
Resistance R = 0.1
Mass of ring m = 2aAd 0.1
Moment of inertia = M = 0.1
Magnetic flux through ring at angle  = Ba2sin  0.1
Instantaneous emf = = Ba2 = Ba2cos t = Bacos  
Induced current = I = Ba2cos /R
Torque opposing motion = (Ba2cos  0.1
Work done in small = 0.1
Average torque = (work done in 2 revolution)/2
= 
This equals so that = 0.2
0.2
0.2
0.2
ln 2 = 0.2

T = =
= 1.10(2) 106 s =306 hr = 12 days 18 hr
0.2
2.0



Theoretical Problem 2


(a) A cathode ray tube (CRT), consisting of an electron gun and a screen, is placed within a uniform constant magnetic field of magnitude B such that the magnetic field is parallel to the beam axis of the gun, as shown in figure 2.3.

The electron beam emerges from the anode of the electron gun on the axis, but with a divergence of up to 5 from the axis, as illustrated in figure 2.4. In general a diffuse spot is produced on the screen, but for certain values of the magnetic field a sharply focused spot is obtained.

By considering the motion of an electron initially moving at an angle  (where   5) to the axis as it leaves the electron gun, and considering the components of its motion parallel and perpendicular to the axis, derive an expression for the charge to mass ratio e/m for the electron in terms of the following quantities:

the smallest magnetic field for which a focused spot is obtained,
the accelerating potential difference across the electron gun V
(note that V < 2 kV),
and D, the distance between the anode and the screen.

Write your expression in the box provided in section 2a of the answer sheet.

(b) Consider another method of evaluating the charge to mass ratio of the electron. The arrangement is shown from a side view and in plan view (from above) in figure 2.5, with the direction of the magnetic field marked B. Within this uniform magnetic field B are placed two brass circular plates of radius  which are separated by a very small distance t. A potential difference V is maintained between them. The plates are mutually parallel and co-axial, however their axis is perpendicular to the magnetic field. A photographic film, covers the inside of the curved surface of a cylinder of radius  + s, which is held co-axial with the plates. In other words, the film is at a radial distance s from the edges of the plates. The entire arrangement is placed in vacuo. Note that t is very much smaller than both s and .

A point source of  particles, which emits the  particles uniformly in all directions with a range of velocities, is placed between the centres of the plates, and the same piece of film is exposed under three different conditions:

firstly with B = 0, and V = 0,
secondly with B= B0, and V = V0, and
thirdly with B = -B0, and V = -V0;

where V0 and B0 are positive constants. Please note that the upper plate is positively charged when V>0 (negative when V0 (in the opposite direction when B