8 campers

4 Balls on a String

3 weight with 2 strings

q2 girl on pulley

  2.     (a)     (pulley wheel) at rest / in equilibrium / acceleration is zero                                B1

(b)     (i)      500 N force down and general shape correct                                            B1

angles correct (one angle labelled correctly)                                             B1

T1 and T2 directions labelled correctly                                                     B1

(ii)     Formulae correct (resolving or sin rule) / scale diagram drawn correctly

with scale given                                                                                     B1

T1 = 674 (N) allow 650 to 700 for scale diagram                                      A1

T2 = 766 (N) allow 740 to 790 for scale diagram                                      A1

q1 masses on pulley

1.(a)  (i)      1.     mass = 360 / 9.8 36.7 (kg)                                                               B1

        (allow 2sf)

(ii)     2.     density = mass / volume                                                                  C1

                   = 36.7 / 4.7 ´ 10–3

                   = 7.8 ´ 10–3                                                                         A1

          unit kg m–3                                                                                   B1

(ii)     right angled triangle with an additional correct angle marked                   M1

set of correct force labels and correct arrows                                            A1

algebra shown or scale given                                                                   C1

tension = 270 (N) or value in the range 255 to 285 (N)                             A1

  (b)   (i)      tension is a vector / has magnitude and direction                                      B1

direction involved in addition / the tensions or ropes act in
different directions                                                                                 B1

(ii)     sum =270 sin37 + 360 sin53                                                                   B1

       =162.5 + 287.5                                                                                B1

(or one mark each for values of 162.5 and 287.5 seen) = 450 (N)              A0

q3 Child on swing

  3.     (a)     weight  = 28 × 9.8 / mg                                                                                    C1
            = 270 (N)      (274.4)                                                                          A1
(using g = 10 then –1)

(b)     a completed triangle drawn with correct orientation                                           B1
at least two labels for triangle with correct directions given                                B1

          calculation:                                     scale diagram:
force P / weight = tan 35                  scale given                                                C1
force P =192 (N)                             185 to 200 (N)                                          A1

  (c)    tension is greater                                                                                             B1
(reference to triangle) tension force would be greater (longer) as the holding
force P would be larger (longer) for greater angle / larger
value needed so vertical component still balances the weight                              B1

Graph of Spring

Resonance expt oscill

Collapsing Tower

Resolving Forces

1. A weight of 50.0 N is suspended from a beam by a string. Calculate the horizontal force must be applied to the weight to keep the string at an angle of 40° to the vertical. 65.3N  Calculate the tension in the string. 
2. A weight of 20 N rests on a plane inclined at 40° to the horizontal. Calculate the components of the weight parallel to the plane and perpendicular to the plane.12.86N, 15.32N
3. A weight of 4.33 N is suspended by a string fastened at its upper end. A horizontal force is applied to the weight so that the string makes an angle of 30⁰ with the vertical. Calculate the force and the tension in the string.2.50N, 5.00N
4. A 10 N weight rests on a smooth inclined plane. A force of 5N parallel to the plane is needed to prevent the weight slipping down the plane. Calculate the reaction of the plane and the angle the plane makes with the horizontal.8.66N ,30⁰
5. A garden roller of weight 800 N is being pulled along by a force of 200 N at 40° to the ground. Calculate (a) the force pulling the roller forwards, 153.2N(b) the vertical force exerted by the roller on the ground. 671.4 N
6. Calculate the forces (a) 153.2N and (b) 928.6N in the previous problem if the roller is being pushed instead of being pulled.

Vector sheet #1

Vectors worksheet

Solutions

1.         6.1N, 5.1N                    2.         -1.4N, 2.6N                   3.         4.6N, -10.0N

4.         -181.3N, -84.5N             5.         -18.4N, -15.4N

Solutions

1.         -6.1N, -5.1N                  2.         1.1N, -2.8N                   3.         -10.1N, 4.3N

4.         1.5N, 4.8N                    5.         9.1N, 4.1N

prep with boat on a river

Energy of a Photon

1.       (a)     quantum of energy / radiation / packet of energy                                               B1

  (b)   (i)      f = E/h = 5.60 × 10–19 /6.63 × 10–34                                                         C1
f = 8.45 × 1014 (Hz)                                                                               A1

(ii)     1   minimum energy to release an electron from the surface (of the metal) B1

2   5.60 × 10–19 – 4.80 × 10–19 (= 8.0 × 10–20 J)                                        B1

(iii)    8.0 × 10–20 = ½(9.1 × 10–31)v2                                                                M1
giving v = 4.2 × 105 (m s–1)                                                                    A1

  (c)    (i)      Correct selection of: λ = h/p or λ = h/mv                                                 M1
where all symbols are defined                                                                 A1

(ii)     λ = 6.6 × 10–34 /(9.1 × 10–31 × 4.2 × 105)                                                  C1
λ = 1.7 × 10–9 (m)                                                                                  A1

[11]

Photoelectric effect

PHOTOELECTRIC EFFECT

Speed of light (c) = 3x10⁸ ms^-1   

Planck’s constant (h) = 6.63 x 10^-34 Js

Mass of an electron = 9x10^-31 kg

hf = f + ½ mv²

wavelength = c / frequency

1. What is the photoelectric effect?

Release of photelectrons from the surface of metal upon illumination

2. What property must the incoming radiation have to cause photoelectric emission from a surface?

Above threshold value

3. What property of the incoming radiation determines the energy of the emitted photoelectrons?

frequency

4. What property of the incoming radiation determines the number of photoelectrons emitted per second?

intensity

5. Radiation of wavelength 120 nm falls on a zinc plate. The threshold wavelength for this surface is 200 nm.

(a)  what is the energy of a quantum of the incident radiation?

E = hf

(b)  what is the energy required to give photoelectric emission?

9.945e-19

(c)   what is the energy of the emitted photoelectrons?

1.6575e-18 - 9.945e-19

6.1e-19

6. When radiation causes photoelectric emission the electron emission is instantaneous - it occurs immediately when radiation falls on the surface. What does this tell you about the incident radiation?

Travels at the speed of light

TAP 502-2: Photoelectric effect questions

hf = f + (1/2) mv² and hf = f + eV_s

e = 1.60 x 10^-19 C,

h = 6.63 x 10^-34 J s,

mass of electron = 9.11 x 10^-31 kg

1          The work function for lithium is 4.6 x 10^-19 J.

(a)        Calculate the lowest frequency of light that will cause photoelectric emission.

(b)        What is the maximum energy of the electrons emitted when light of 7.3 x 10¹⁴ Hz is used?

2          Complete the table.

Metal	Work Function /eV	Work Function /J	Frequency used /Hz	Maximum KE of Ejected electrons /J
Sodium	2.28		6 x 1014
Potassium		3.68 x 10-19		0.32 x 10-19
Lithium	2.9		1 x 1015
Aluminium	4.1			0.35 x 10-19
Zinc	4.3			1.12 x 10-19
Copper		7.36 x 10-19	1 x 1015

3          The stopping potential when a frequency of 1.61 x 10¹⁵ Hz is shone on a metal is 3 V.

(a)        What is energy transferred by each photon?

(b)        Calculate the work function of the metal.

(c)        What is the maximum speed of the ejected electrons?

4          Selenium has a work function of 5.11 eV. What frequency of light would just eject electrons? (The threshold frequency is when the max KE of the ejected electrons is zero)

5          A frequency of 2.4 x 10¹⁵ Hz is used on magnesium with work function of 3.7 eV.

(a)        What is energy transferred by each photon?

(b)        Calculate the maximum KE of the ejected electrons.

(c)        The maximum speed of the electrons.

(d)        The stopping potential for the electrons.

Answers and worked solutions

1(a)     hf = f

hf = 4.60 x 10^-19

f = 4.60 x 10^-19 / 6.63 x 10^-34 = 6.94 x 10¹⁴ Hz

 (b)      hf = f + (1/2) mv².   (6.63 x 10^-34 x 7.30 x 10¹⁴) = 4.60 x 10^-19 + (1/2) mv²

4.84 x 10^-19 - 4.60 x 10^-19 = (1/2) mv² = 0.24 x 10^-19 J

2

Metal	Work Function / eV	Work Function / J	Frequency used / Hz	Maximum KE of ejected electrons / J
Sodium	2.28	3.65 x 10-19	6 x 1014	0.35 x 10-19
Potassium	2.30	3.68 x 10-19	6 x 1014	0.32 x 10-19
Lithium	2.90	4.64 x 10-19	1. x 1015	1.99 x 10-19
Aluminium	4.10	6.56 x 10-19	1.04 x 1015	0.35 x 10-19
Zinc	4.30	6.88 x 10-19	1.2 x 1015	1.12 x 10-19
Copper	4.60	7.36 x 10-19	1 x 1015	0

For copper 1 x 10¹⁵ Hz is below the threshold frequency so no electrons are ejected.

3)

(a)       1.07 x 10^-18 J

(b)       hf = f + eV_s, so f = hf - eV_{s, so} f = 1.07 x 10^-18 – (1.6 x 10^-19 x 3) = 5.9 x 10^-19 J

(c)       eV_s = (1/2) mv² so (1.60 x 10^-19 x 3) = 0.5 x 9.11 x 10^-31 x v² 

so v² =1.04 x 10¹² and v = 1.02 x 10⁶ m s^-1

4          1.2 x 10¹⁵ Hz

5

(a)       1.6 x 10^-18 J

(b)       (1/2) mv² = 1.0 x 10^-18 J

(c)       v² = 1.1 x 10¹² so v = 1.1 x 10⁶ m s^-1

 (d)      (not in spec) eV_s = (1/2) mv² so eV_s = 1.00 x 10^-18 and V_s = 0.63 V

Questions on E Fileds

magnetism questions