Gpe of Block in water

10 Sledge

Toy Frog

1 Bullet

6 aircraft power

1 Water Skier

5 mass on slope

Question on Principle of Moments

3. (a) Principle of moments

In equilibrium

sum of clockwise moment (about any point) is equal to sum of

anticlockwise moment (about that point)

2

(b)(i) Weight

Use of “width x thickness x length”

Use of “density = mass”

volume

Correct value 3

V = 1.2 × 0.6 × 200 (cm3) = 144 (cm3)

Using ρ = m/v

m = 8 (g cm-3) × 144 (cm-3) = 1152 g

Weight = mg = 1152 × 10-3 (kg) × 9.81 ( m s-2) = 11.3 (N) / 12

(N)

(ii) Force F

Correct substitution into correct formula

Correct value with correct unit

F × 60 (cm) = 11.3 (N) × 40 (cm) / 12 (N) × 40 (cm) / 11(N) × 40

(cm)

= 7.5 N / 8 N / 7.3 N

2

(iii) Force R

18.3 N / 18.8 N / 20 N

1

(iv) Sketch graph

Any line upwards

Correct shape for F [concave shaped curve]

2

Moments of Forearm

Mark Scheme Capacitor questions

1.       (a)     (i)      Cp = 2 + 4 = 6 μF                                                                                  A1

  (ii)   1/C = 1/2 + ¼                                                                                        C1
Cs = 4/3 =1.33 μF                                                                                 A1

  (b)   (i)      6.0 V                                                                                                    A1

  (ii)   Q = CpV                                                                                                C1
= 6 × 6 = 36 μC                                                                                    A1

  (c)   E = ½ CsV2                                                                                                    C1
= 24 × 10–6                                                                                                    A1

  (d)   (i)      The capacitors discharge through the voltmeter.                                   B1

  (ii)   V = V0e–t/CR
1/4 =e–t/(6×12)                                                                                        C1
ln 4 = t / 72                                                                                           C1
t = 72 ln 4 ≈ 100 s                                                                                A1

[12]

 

  2.     (a)     Qo = CV = 1.2 × 10–11 × 5.0 × 103; = 6.0 × 10–8; C (3)                                    3

  (b)   (i)      RC = 1.2 × 1015 × 1.2 × 10–11 or = 1.44 × 104 (s) (1)                              1

(ii)     I = V/R = 5000/1.2 × 1015 or = 4.16 × 10–12 (A) (1)                               1

  (iii)  t = Qo/I; = 6 × 10–8 / 4.16 × 10–12 = 1.44 × 104 (s)                                  2

  (iv)  Q = Qoe–1; Q = 0.37Qo so Q lost = 0.63Qo                                              2

  (c)   (i)      capacitors in parallel come to same voltage (1)
so Q stored α C of capacitor (1)
capacitors in ratio 103 so only 10–3 Qo left on football (1)                      3

  (ii)   V = Q/C = 6.0 × 10–8 /1.2 × 10–8 or 6.0 × 10–11 /1.2 × 10–11 or only 10–3
Q left so 10–3 V left; = 5.0 (V)                                                                2

[14]

 

  3.     (a)     (i)      Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii)     parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1)                                                       4

  (b)   (i)      T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h                                     2

(ii)     ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J)                                       2

  (iii)  4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s)    2

(iv)    P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW            ecf b(ii) and (iii)                 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]

 

  4.     (a)     29; 34                                                                                                               2

(b)     λ = 0.693/T = 0.693/(120 × 3.2 × 107) = (1.8 × 10–10 s–1) accept ln 2              1

  (c)   (i)      Q = CV = 1.2 × 10–12 × 90; evidence of calculation (= 1.1 × 10–10 C)    2

  (ii)   n = Q/e = 1.1 × 10–10/1.6 × 10–19; = 6.9 × 108 allow sig. fig. variations  2

(iii)    A = λN; N = 6.9 × 108/1.8 × 10–10; = 3.8 × 1018 using 7.0 gives 3.9       3

  (iv)  1 y is less than 1% of 120 y so expect to be within 1%/
using e–λt gives exactly 1% fall/ problem of random emission
or other relevant statement                                                                      1

[11]

 

  5.     (a)     (i)

capacitor	capacitance / µF	charge / µC	p.d. / V	energy / µJ
X	5	30	= Q/C = 6 (V) (1)	= ½ CV 2(1) = ½ × 5 × 62 = 90 (1)
Y	25	= CV = 25 × 6 =150 (µC) (1)	= 6 (V) (1)	= 450 (1)
Z	10	30 + 150 = 180 (µC) (1)	= Q/C = 180/10 = 18 (V) (1)	= 1620 (1)

         Each box correctly calculated scores (1) + (1) for ½ CV2                        9

  (ii)   1   18 V + 6 V = 24 (V) (1)

2   180 (µC) (1)

3   180 / 24 = 7.5 (1)

4   90 + 450 + 1620 = 2160 (µJ) (1)                                                        4

  (b)   (i)      Kirchhoff’s second law OR conservation of energy (1)                          1

(ii)     Kirchhoff’s first law OR conservation of charge (1)                               1

  (c)   (i)      time constant = CR (1)
= 7.5 × 10–6 × 200 000 = 1.5 (s) (1)                                                        2

  (ii)    (1)
Q/Qo = e–4 = 0.0183 (1)                                                                          2

[19]

 

  6.     (a)     (i)      5.0 (V) (1)                                                                                               1

(ii)     10.0 (V) (1)                                                                                             1

(b)     (i)      Q = CV;= 1.0 × 10–3 (C) (2)                                                                    2

(ii)     The total capacitance of each circuit is the same (namely 100 μF); (1)
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks                 2

  (c)   (i)      A1 will give the same reading as A2; because the two ammeters are (1)
connected in series /AW (1)                                                                    2
answer only in terms of exponential decrease for a maximum of 1 mark

  (ii)   A4 will show the same reading as A2 at all times; (1)
A3 will show half the reading of A2 initially;
and at all subsequent times (2)                                                                3

[11]

 

  7.     (i)      C = Q/V or gradient of graph / = 24 μC/3V; = 8.0 (μF)                                     2

(ii)     E = ½ CV2 / = ½ × 8 × 32; = 36 (μJ) ecf a(i)                                                      2
or ½ QV / = ½ × 24 × 3; = 36 (μJ)

  (iii)  T = RC = (0.04); R = 0.04/8.0μ = 5.0 × 103 (Ω) ecf a(i)                                    2

(iv)    idea of exponential/constant ratio in equal times; which is independent of
initial value/AW or argued mathematically in terms of Q/Qo = e–t/RC
give 1 mark for statement that time depends only on time constant/RC            2

[8]

 

  8.     (i)      Cp = C + C = 6 μF; 1/Cs = 1/2C + 1/C; = 3/2C giving Cs = 2C/3 = (2 μF)          3

(ii)     2 sets of (3 in series) in parallel/ 3 sets of (2 in parallel) in series                     2

[5]

 

 

Car Safety

Momentum and Safety

1 The car takes 4 seconds to stop. Calculate the force exerted by the brakes.

 2500 N

                   

2 The same car is driven into a wall. It is traveling at 20 m/s. The car comes to a halt in 0.006 s. Calculate the force exerted by the wall on the car.

F =  1 666 667N = 1.67MN

 

Compare this with the answer to the previous question. Very much larger

What affect does the time taken to stop have on the force applied to the car? The longer the time to stop the lower the acceleration the less force exerted

3  A human head has a mass of 4.5 kg.

If the person is not wearing a seat belt and the car what happens to the speed of the head? It will hit the windscreen at 15m/s  What is the speed of the head after it has hit the windscreen? 0m/s The head stops in 0.002 s. calculate the force exerted on the head.

F  =  33750 N 

 If the car had air bags fitted the head is brought to a halt in 0.15s. Calculate the force exerted on the head now. F =   450 N

  

4 A driver has a mass of 65kg she is not wearing a seat belt. In a crash her car stops and she hits the steering wheel which stops her in 0.05s. Calculate the force exerted on her by the steering wheel. F = 45 500 N

Had she been wearing a seat belt she would have come to a halt in 0.2s.

Calculate the force exerted on her body by the seat belt. . F = 11 375 N

 

 5 A car has a mass of 750 kg and a velocity of 15m/s. (a)Calculate its KE. ½ x 750 x 15² = 84375 J (b) What is the work done by the brakes to stop the car 84375 J The brakes exert a maximum force of 5600 N. (c) Calculate the distance travelled before the car stops. Work = F x d, so, d = work / F = 84375/ 5600 = 15m(d) Repeat this for a speed of 30m/s KE = ½ x 750 x 30² = 337 500 J               Work = F x d, so, d = work / F = 337 500/ 5600 = 60.3m

 

 

 

 

 

 

 

General Questions on capacitance

Capacitors

 

1 A capacitor has a charge of 20 μC (microcoulomb) when a p.d. of 200 V is applied to it. Calculate the capacitance of the capacitor. 1x10^-7F

What is the charge on the capacitor if a battery of 40 V is connected?

          4 x10^-6C
2 Define capacitance, microfarad. Calculate the p.d. across a 2 μF capacitor if it has a charge of 80 μC. 40V Calculate the new p.d. if the capacitor is then connected to an uncharged capacitor of 4 μF. 13.3V What is the charge on each capacitor in this case? 2.7 x10^-5 C, 5.3 x10^-5 C

3 Calculate the combined capacitance of (i) 2 μF and 3 μF capacitor in series, 1.2 μF (ii) a 4 μF capacitor in series with a parallel arrangement of a 3 μF and 2 μF capacitor. 2.2 μF Prove from first principles the formula 4 for the combined capacitance of two capacitors in series and in parallel.

4 A capacitor of 2 μF is charged by a 100 V battery. Calculate the energy in the capacitor. 1 x10^-2 J If the capacitor is disconnected from the battery and then connected to a 6 μF uncharged capacitor, find the new energy in each capacitor 6.3 x10^-4 J, 1.9 x10^-3 J. Account for the loss in energy which has occurred. Heat lost as current passes through external in circuit, (resistance of wires)

5 A capacitor of 4 μF is (i) in parallel, 4.8 x10^-4 C 7.2 x10^-4 C (ii) in series with a 6 μF capacitor, 2.9 x10^-4 C and a battery of 120 V is connected across the arrangement in each case Calculate the charge on each capacitor, and the total energy of the capacitors in both cases. Parallel 72mJ Series 17.2mJ

6 Two capacitors of 25 μF and 100 μF respectively are joined in series with a d.c. supply of 6.0 V. Fig 3 (i) What is the charge on each capacitor and the p.d. across each? 1.2 x 10^-4 C, 4.8V, 1.2V
The supply is now disconnected without affecting the charge on each capacitor. Their two positive plates, and their two negative plates, are then connected together Fig. 3 (ii). Calculate (i) the common p.d. of the capacitors, 1.92 V(ii) the loss in energy of the two capacitors. 1.3 x10^-4 J. How is this loss of energy accounted for?
7 A 100 V supply is connected to a 4 μF capacitor in series with a 2 MΩ (2 million ohms) resistor. Find (i) the current I at the instant of switching on the supply 5 x10^-5A, (ii) the final charge on the capacitor4 x10^-4C, (iii) the time taken to charge the capacitor assuming the mean value of the current during flow of charge is I/2. 16s (I do know of the approximation of 5RC, but you have to use the information in the question)
8 A 25 μF capacitor, previously charged by a p.d. of 10 V, is discharged through a 2 MΩ(2 x 10⁶Ω) resistor. What is: (i) the initial charge on the capacitor 2.5 x10^-4C  (ii) the initial current? 5 x10^-6 A