Pre-IGCSE Problem set 1

8/17/2019 Pre-IGCSE Problem set 1

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Solved Problems – IGCSE Physics Mr. Anas Elgaud

Don’t Panic

10. On the Earth the acceleration of free fall is about 10 m/s2. On the moon the

acceleration of free fall is about 1.6 m/s2 (which is about one-sixth the

acceleration on the Earth).

A man weighs 800 N on the Earth. Give his mass measured on the moon, and

the weight measured on the Moon

10. Weight on the Earth = 800 N. W = mg, therefore m = W/g = 80 kg. Mass on

the Earth = Mass on the Moon = 80 kg;

Weight measured on the moon = m x g on moon = 128 N

11. A mass of 0.5 kg extends a spring by 10 cm. When an unknown mass M, is

hung on the spring, the extension is 15 cm. What is the value of M?

11. “A mass of 0.5 kg extends a spring by 10 cm”, from this use F = kx, solve

for k. k = F / x. My F here is the force of weight (W =mg=5N)

Then, k = 5N/0.1m=50N/m.

Now I have my k, I can use it with an extension of 15 cm, to find the requiredweight (from which I find the required mass)

F = k x = 50 N/m x 0.15 m = 7.5 N; Mass of weight 7.5 N is 0.75 kg

12. Which see-saw in the diagram below is balanced? In which direction will

the unbalanced see-saw tip?

12. To solve any problems containing moments, we start by finding the

moments in both clockwise and counterclockwise directions, and then we

compare them.

A. CW = 10 N x 2.5 m = 25 Nm; CCW = 150 N x 2.5 m = 375 Nm

CCW moment is larger than CW moment. Therefore it tips in CCW direction

B. CW = 200 N x 2 m = 400 Nm; CCW = 150 N x 2.5 m = 375 Nm

CW moment is larger than CCW. Therefore it tips in CCW direction

C. CW = 300 Nm. CCW = 300 Nm. CW = CCW. Therefore it is balanced.

D. CW = 350Nm. CCW = 375 Nm. Tipping in CCW


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13. A dog pulls a sledge across a flat snowfield for a distance of 500 m.

Calculate the work done if the dog exerts a steady force of 120 N on the

sledge.

13. Work done = force x distance moved = 120 N x 500 m = 60 KJ (Note: work

done = energy transferred)

14. An electrical appliance has a power input of 500 W. If it transfers energy

at a rate of 350 W, calculate its efficiency. What happens to the ‘lost’ energy?

14. Efficiency = 70%; ‘Lost’ energy is not really lost, it’s transferred to non-

useful heat.

15. A man lifts a weight of 60 N through a vertical distance of 1.2m. How

much work is done? If the man takes 2 minutes to lift 16 boxes each of weight

60 N vertically through 1.2 m, at what power is he working?

15. Work done = energy transferred = force x distance = mg x h = 60 N x 1.2 m= 72 Joules.

For 16 boxes, the work (energy) is 72 J x 16 = 1152 J. Time is 120 seconds.

Hence, power = 1152 J per 120 seconds = 1152J/120s=9.6 Watt

16. A lift raises a load of 9000 N to a height of 20 m in 15 seconds. Calculate

the work done, and the power of the lift

16. Work = 180Kilojoules. Power= 12Kilowatt

17. A 3 kg mass falls with its terminal velocity. Which of the combinations A to

E gives its weight, the air resistance and the resultant force acting on it?

Answer is D, the body feels weightless in free-fall at terminal velocity because

the forces balance on it. So its velocity is not increasing, but it is still moving

(at constant velocity, i.e. the terminal velocity)


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18. Which one of the following statements is not true?

A Pressure is the force acting on unit area.

B Pressure is calculated from force/area.

C The SI unit of pressure is the pascal (Pa) which equals 1 newton per

square metre (1 N/m2).

D

The greater the area over which a force acts the greater is thepressure.

E Force = pressure × area.

Answer: This is an easy one; clearly it is D since P = F / A, if we increase A, P

decreases (Logic 101)

19. A woman cycles, along a level road, a distance of 2.4 km in 10 minutes.

What is the woman’s average speed? The woman does work during her cycle

ride. Against what force has this work been done?

19. Average speed = Total distance / time taken = 2400 m / 600 s = 4 m/s;

work has to be done against friction between the wheels and the road.

20. A bus starts from rest and accelerates smoothly. After 10 s the bus reaches

a speed of 8 m/s, a) find the acceleration of the bus. b) If the bus continues to

travel at 8 m/s and then decelerates smoothly as it approaches a bus stop,

and the deceleration is 8 m/s2, find the time over which the bus decelerates

before it comes to rest

20. a) a = (v-u) / t = (8m/s – 0m/s) / 10 s = 0.8 m/s2.

b) a now = - 8 m/s2. Since the bus is coming to a stop (decelerating).

a = (v-u) / t; HENCE, t = (v-u) / a = (0m/s – 8m/s) / (-8 m/s2) = 1 s

21. In the following page, a graph showing at the top the distance-time graph

for a girl’s bicycle ride and the bottom you are given the axes for the

corresponding speed-time graph.

(i) What is happening to the distance from the starting point between A and

B?

(ii) What can you say about the speed of the bicycle?

(iii) On the speed-time axes on the bottom graph, draw a thick line that could

show the speed during AB.

(iv) Find the speed during BC and CD; and draw the lines for the speed -time

portions accordingly


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22. Define Brownian motion, with a scribble of a graph. SB-P72

23. Describe Bimetallic strip with a sketch. SB- P82

24. Write formulas of Specific heat capacity SB- P88, and specific latent heat

(SB-P92) – Can you describe experiments to measure both c and L? If it’s not

any trouble, of course.

Comment : These are some general revision questions that should helpconsolidate your understanding of our current coursework in Physics. If you

have any points to be clarified reach me on our deserted fb group.

Mr. Anas Elgaud

IGCSE Physics

4/4/2016

Pre-IGCSE Problem set 1

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