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

Triangle of Forces (Past questions)

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

[12]

 

  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

[7]

 

  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

[8]

 

  4.     (a)     (i)      1 Horizontal component = 24cos30                                                      C1

                                     = 21    (20.8) (N)                                              A1

2. vertical component = 24sin30

                                 = 12        (12.0) (N)                                             A1

(ii)     vertical force = 65 + 12                                                                        M1

                             = 77                                                                                  A0

(iii)    horizontal force = 20.8 (note ecf for 20.8 component)

resultant = [(77)2 + (20.8)2]1/2                                                               C1

                          = 80 (79. 8) (N)                                                          A1

(or by vector triangle need correct labels and arrows for C1 mark)

(iv)    80 (79.8)(N) / equal to (iii)  allow ecf                                                   B1

the resultant force needs to be zero or forces need

to balance above value to give no acceleration or constant velocity     B1

  (b)   (i)      P = F / A                                                                                               C1

    = 77 / 4.2 ´ 10–3

    = 18000 (18333) (Pa)                                                                       A1

(ii)     more / increases

downward / vertical component (of P) will be greater                           B1

(for larger angles)

[11]

 

 

Triangle of forces

Triangle of Forces

1. The ends of a string 9 m long are tied to two points 6m apart on a horizontal beam. A 10 N weight is attached to the string 4 m from one end. Find the tension in each part of the string.
2. A string 7 m long is attached to two points 5 m apart on a horizontal bar. A 5 N weight is attached to the string 3 m from one end. Find the tension in each part of the string.
3. Two forces of 10 N and 16 N act at 60° to each other. What is the magnitude and direction of the force needed to balance them?
4. Three forces of magnitude 4 N, 6 N and 5 N are in equilibrium. Find the angles between them.
5. A 12 N weight is supported by two strings, their tensions being 8.2 N and 6.4 N. Find the angle each string makes with the horizontal.

Some calcutations on forces and moving objects

I have made a number of mistakes. Your task is to copy and correct my answers as you go.

Laws of Motion

(Use g = 9.8Nkg^-1)

 

1. Find the acceleration of a body of mass 10kg when it is subjected to a horizontal force of 100 N if it (a) can move along a smooth horizontal surface, Ignoring Friction a = F/m = 100/10 = 10ms^-2 (b) can move along a horizontal surface which produces a frictional force of 80 N. Resultant Force = 100 – 80 = 20N, using a = F/m = 80/10 = 8 ms^-2

 

2. A rocket of mass 800 000 kg has motors giving a thrust of 9 800000 N. Find the acceleration at lift off Ignoring air resistance a = F/m = 8 x 10⁵ / 9.8 x 10⁶ = 8.1632653061224489795918367346939e-2

 

3. A force of 100 N acts on a mass of 1 kg. What is the acceleration? Ignoring friction 100ms^-1

 

4. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface. What is its velocity after 5s? Ignoring friction  a = (v-u)/t = F/m. Rearranging v – u = Ft/m As u = 0 v = Ft/m = 5 x 5/2 =12.5 ms^-2

 

5. A mass of 10 kg acquires a velocity of 20 ms^-1 from rest in 4 s. What force is required?  Ignoring friction F=m a rearranging F= m (v-u)/t = 10(4-0)/20 = 2N

 

 

6. A car of mass 600 kg travelling at 72 km h is brought to rest in 54 m after the driver sees an obstruction ahead. If the distance travelled after the driver applies the brakes is 40 m find the driver’s reaction time Distance travelled in time taken for driver to react = 72 – 40 = 32m.  v= x/t so t = x/v = 32/72 = .44 s and the braking force v² = u² + 2as  rearranging a= (v² – u²) / 2s. As F = ma then F = m (v²-u²)/2s = 600(0 – 20²)/2 x 40 = 3000N

 

7. A mass of 2kg projected along a flat surface with a velocity of 15 m s^-1 comes to rest after travelling 30 m. What is the frictional force?

 

 v² = u² + 2as  rearranging a= (v² – u²) / 2s. As F = ma then F = m (v²-u²)/2s = 2 x 15/60 = 0.5 N

8. A Mini of mass 576 kg can accelerate from rest to 72km h-1 in 20 s. If the acceleration is assumed uniform find this acceleration and the tractive force in Newtons needed to produce it.

F=ma = m (v-u)/t = 567 (20 – 0)/20 = 567N

 

9. A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 km h-1 Find (a) the deceleration, v² = u² + 2as  rearranging a= (v² – u²) / 2s. = 1082 / 2 x 72 = 7.5 ms^-2(b) the frictional force between the tyres and the road in Newtons, F=ma = 576 x 7.5 = 4328 N

 

10. The first-stage rocket motors of the Apollo spacecraft produce a thrust of 3.3 x 10⁷ N and the complete spacecraft has a mass of 2.7 x 10⁶ kg. Find (a) the resultant force accelerating the spacecraft,

 

W= mg  = (2.7 x 10⁶) 9.81 = 2.6 x 10⁷. Resultant force = Thrust – weight = (3.3 x 10⁷ - 2.6 x 10⁷ )= 6.5 x 10⁶ N

(b) the initial acceleration, a = F/m = 6.5 x 10⁶ / 3.3 x10⁷ = 0.197 ms-²

(c) the time for it to rise through a distance equal to its own height as it ‘lifts off if its height is 111 m and the average acceleration during this time is 2.5 ms^-2
s = ut + ½ at2 As u = 0 then s = ½ at2 rearranging t= √2s/a = √222/2.5 = 5.96 s

11. A boy of mass 50kg stands in a lift. What will he ‘weigh’ in Newtons if the lift accelerates at 0.50 ms^-2 (a) upwards, Force of lift on boy due to acceleration = ma = 50 x 0.5 25N Force of lift on boy due to his weight = 50 x 9.81 = 490.5N total force exerted on lift by boy = 25 + 590.5 = 515.5N  (b) downwards? 465.5N

 

Resolving Forces

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. T cos40 = 50.0 T = 50/cos40 = 65.3N  Calculate the tension in the string. 42.0 N
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.

Answer to capacitor question

(a)

V across 200k resistor = IR

V= (2x10⁵) (2x 10^-5) = 4V

V_C = E - V_R = 6-4 = 2V

(b)

Q=VC

Q = (2.0) (560 x 10^-6) = 1.12 x 10^-3 C

 

(c)

W = ½ CV²

W = ½ (560 x10^-6) (2)² = 1.12 x 10^-3 J

 

(d)

 W = Emf x charge

W = 6V x 1.12 x 10^-3 C = 6.72 x 10^-3 J

 

(e) Work is done transferring the charge through the external resistance. This is transferred to heat energy in the resistor

y12 Sankey Diagrams

Constructing Sankey Diagrams








1. An energy efficient light bulb is rated at 20W. It produces 5W of light. Calculate its efficiency and draw a Sankey Diagram to scale.



2. Paula transfers 40 000J of chemical energy during a race. She transfers 32 000J of heat energy to the surroundings during the race. Calculate her efficiency and draw a Sankey diagram to scale.



3. Bradley does 1600J of work turning the pedals on his bike. 1577.6J is transferred to the rear sprockets. How much heat is lost and what is the efficiency of Bradley’s bike. Draw a Sankey diagram of Bradley’s chain



4. The coal in Thomas’ boiler contains 40 kJ of energy. He loses 25.76J as puffs along a branch line. How efficient is Thomas and draw a Sankey diagram.



5. The fuel in Diesel’s tank contains 60kJ of chemical energy. He does 20.1kJ of work on the mainline. Calculate Diesel’s efficiency and draw a Sankey diagram.



6. Jeremy is the proud owner of an electric car. The electric battery in Jeremy’s car can develop 3kW. If the engine develops 2.658kW what is its efficiency.



7. Jeremy dreams of a Ferrari Enzo which can develop a maximum of 700 bhp. (1 bhp = 750 Watts). Sadly for Jeremy petrol cars are not very efficient. Typically, only about 30% of the energy that is available from the combustion of the petrol actually ends up overcoming friction to move the car forwards. Of the 70% of energy is that is not usefully converted, 55% may heat the cooling water that surrounds the engine block whilst 15% may be in the hot exhaust gases. To make car engines more efficient the fuel has to burn at a higher temperature and the exhaust must be kept cooler. Draw a Sankey diagram of Jeremy’s dinosaur.

Y12 Newton's Second Law

Newton’s Second Law of Motion
(Use g = 9.8 N kg^-1)

Worked example

Find the acceleration of a body of mass 10kg when it is subjected to a horizontal force of 100 N if it (a) can move along a smooth horizontal surface, (b) can move along a horizontal surface which produces a frictional force of 80N.
(a) F=ma      F force in N, mass in kg, a acceleration in ms^-2
Rearranging  a = F/m        a = 100 / 10  = 10 ms^-2

(b) The resultant force = 100 N – 80 N = 20 N

F=ma rearranging a = F/m
 a = 20 /10   a = = 2ms^-2  

1. A force of 100 N acts on a mass of 1 kg. What is the acceleration?
2. A mass of 10kg acquires a velocity of 20 m s from rest in 4 s. Calculate the force is required.
3. A rocket of mass 800 000 kg has motors giving a thrust of 9 800 000 N. Calculate the acceleration at lift off.
4. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface. Calculate its velocity after 5 s?
5. A car of mass 600 kg travelling at 72km h^-1 is brought to rest in 54 m after the driver sees an obstruction ahead. If the distance travelled after the driver applies the brakes is 40m calculate the driver’s reaction time and the braking force.
6. A mass of 2kg projected along a flat surface with a velocity of 15ms^-1 comes to rest after travelling 30 m. Calculate the frictional force.
7. A Mini of mass 576 kg can accelerate from rest to 72 km h^-1 in 20 s. If the acceleration is assumed uniform calculate this acceleration and the tractive force in Newtons needed to produce it.
8. A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 kmh^-1. Calculate (a) the deceleration, (b) the frictional force between the tyres and the road in Newtons.
9. The first-stage rocket motors of the Apollo spacecraft produce a thrust of 3.3 x 10⁷ N and the complete spacecraft has a mass of 2.7 x 10⁶ kg. Calculate (a) the resultant force accelerating the spacecraft, (b) the initial acceleration, (c) the time for it to rise through a distance equal to its own height as it ‘lifts off’ if its height is 111 m and the average acceleration during this time is 2.5 ms^-2
10. A boy of mass 50kg stands in a lift. What will he ‘weigh’ in Newtons if the lift accelerates at 0.50 ms^-2 (a) upwards, (b) downwards?
    
    

 

y12 Force

Force

F = force (Newtons)       m = mass (kg)               a= acceleration (ms-2)

 

 

 

1. A car hits a wall and decelerates at 40 ms^-2.  It exerts a force of 20 000 N on the wall.  Calculate the mass of the car?

 

2. A stone of mass 0.5 kg falls into a sandpit exerting a force of 5N.  Calculate its acceleration?

 

3. A car of mass 1000 kg travelling at 10 ms^-1 stops in 4 s. Calculate a) the average deceleration. b) the braking force.

 

4. An aeroplane accelerates uniformly from rest (0) to 200 ms^-1 in 40 s. Calculate the force it exerted on a pilot whose mass is 90 kg?

 

5. A bullet of mass 20 g travelling with a velocity of 300 ms^-1 hits an earth bank and is brought to rest in 0.2 s Find      a) the deceleration of the bullet. b) the force it exerts on the earth bank.

 

6. A car of mass 750 kg stops uniformly in 5s.  If its initial velocity was 15 ms^-1. Calculate the braking force.

 

7. The valve of a cylinder on an engine containing 12 kg of compressed gas is opened and the cylinder empties in 1min 40 sec. If the gas leaves its chamber at 25 ms^-1.Calculate (a) the acceleration of the gas. (b) the force exerted on the chamber.