SQA National 5 Physics 2017 – Section 1

hello physicists today we’re going to have a look at the esquire’s SQA National 5 Physics 2017 and focus on the multiple choice questions from section one

1. A cyclist is travelling along a straight road. The graph shows how the velocity of the cyclist varies with time.

The kinetic energy of the cyclist is greatest at

  • A    P
  • B    Q
  • C    R
  • D    S
  • E     T

Ans : P

Explanation

So question number one a cyclist is traveling along a straight road the graph shows how the velocity of the cyclist varies with time the kinetic energy of the cyclist is greatest at so we should know that kinetic energy is equal to a half m v squared therefore the greater the speed that the cyclist is moving the greater the kinetic energy the point on the graph with the greatest speed is point p which is answer

Ans : B

Explanation

 

a question two a student has set up a circuit setup as shown rather the reading on ammeter a1 is 5 amps the reading on ammeter a2 is 2 amps the charge passing through the lamp in 30 seconds is okay so the first thing that we need to realize here is that five amps is leaving that power supply there and coming around the circuit but when it comes to the parallel section in the circuit that current splits two amps going along the middle branch through the resistor and five minus two which leaves three amps to pass through the lamp now we can use the formula q equals i times t current times time so three amps times 30 seconds which gives us 90 coulombs answer b

3. A lamp is connected to a constant voltage power supply. The power supply is switched on.
The graph shows how the current in the lamp varies with time.

Ans : A

Explanation

Number three a lamp is connected to a constant voltage power supply the power supply switched on the graph shows how the current in the lamp varies with time which row in the table shows what happens to the current and resistance of the lamp between 0.05 seconds and 0.45 seconds well if we have a look at that there we can see that the current is definitely decreasing if the current is decreasing then the resistance must have increased at the same time answer

4. A circuit is set up as shown.

The purpose of the transistor is to

  • A supply energy to the circuit
  • B decrease the voltage across R1
  • C change electrical energy to kinetic energy
  • D supply energy to the motor
  • E switch on the motor.

Ans E

Explanation

a question four a circuited setup is shown the purpose of the transistor is two well the purpose of any transistor is to be an electronic switch and in this case it’s an electronic switch which is going to switch on the motor

5. Five students each carry out an experiment to determine the specific heat capacity of
copper. The setup used by each student is shown.

  • A student 1
  • B student 2
  • C student 3
  • D student 4
  • E student 5.

Ans : B

Explanation

Number five five students each carry out an experiment to determine the specific heat capacity of copper the setup used by each student is shown the student with the setup that would allow the most accurate value for the specific heat capacity of copper is determined to be well what have we got going on here we have five different setups some of the setups have insulation wrapped around the copper block insulation will stop heat being lost to the surroundings and will stop the heater from heating the air around about the block which means there would be a more efficient transfer of heat into the block so that would improve it so student 2 and student 5 have that additionally if we’re measuring the specific heat capacity of copper and to find out how transferring energy to copper causes this temperature to increase we need to be measuring the temperature of the copper you’ll notice student two has inserted his thermometer into the copper block where student five has left the thermometer outside the thermometer in the block will give us a more accurate reading of the temperature of the copper block so student number two has got the better set up so that gives us answer b

6. The mass of a spacecraft is 1200 kg.
The spacecraft lands on the surface of a planet.
The gravitational field strength on the surface of the planet is 5·0N kg-1
.
The spacecraft rests on three pads. The total area of the three pads is 1·5 m2
.
The pressure exerted by these pads on the surface of the planet is
A 1·2 × 104 Pa
B 9·0 × 103 Pa
C 7·8 × 103 Pa
D 4·0 × 103 Pa
E 8·0 × 102 Pa

Ans : D

Explanation

number six the mass of a spacecraft is 1200 kilograms the spacecraft lands on the surface of a planet the gravitational field strength on the surface of the planet is five newtons per kilogram the spacecraft rests on three pads the total area of the three pads is 1.5 square meters the pressure exerted by these pads on the surface of the planet is so pressure is force over area we need to know what force the spacecraft acts on the planet the force that the spacecraft acts is going to be its weight so w equals m times g its mass 1200 times the gravitational field strength on the planet which is five which gives us six thousand newton’s worth of force the pressure as we said is force over area which in this case is 6000 newtons divided by a total area of the three pads of 1.5 square meters which gives us a pressure of 4 000 pascals or four times ten to five three pascals which is answer d

7. A solid is heated from -15 °C to 60 °C. The temperature change of the solid is

  • A 45 K
  • B 75 K
  • C 258 K
  • D 318 K
  • E 348 K.

Ans : B

Explanation

Question number seven a solid is heated from minus fifteen degrees celsius to sixty degrees celsius the temperature change in the solid is well all we need to do is do 60 minus minus 15 which gets us to 75 15 degrees from minus 15 degrees to zero and then another 60 degrees from 0 up to 60 which gives us a total of 75. the fact that these temperatures are measured in celsius and we want an answer in kelvin is not really relevant here because a change in temperature in kelvin is the same as a change in temperature in celsius we can prove this if we change the initial temperature of 15 celsius in minus 15 celsius and kelvin rather we get 258 kelvin if we change the final temperature of 60 celsius into kelvin we get 333 kelvin the difference in temperature is still 75 kelvin so we didn’t need to do the conversion answer b 75 kelvin

8. A student makes the following statements about waves.
I In a transverse wave, the particles vibrate parallel to the direction of travel of the wave.
II Light waves and water waves are both transverse waves.
III Sound waves are longitudinal waves.
Which of these statements is/are correct?

  • A I only
  • B II only
  • C III only
  • D I and II only
  • E II and III only

Ans : E

Explanation

number eight a student makes the following statements about waves in a transverse wave the particles vibrate parallel to the direction of travel of the wave light waves and water waves are both transverse waves sound waves are long internal waves which of these statements is or are correct well let’s consider the first one first in the transverse wave the particles vibrate parallel to the direction of the wave well here’s a transverse wave and in this one we can see the particles are vibrating in the vertical plane whereas the wave is moving left to right so the particles are not vibrating parallel to the directional wave they are vibrating perpendicular to the direction of the wave so that is incorrect the second statement light waves and waterways are both transverse waves well that is true sound waves are longitudinal waves that is true so both two and three are correct so that gives us answer e

Ans : C

Explanation

number nine the diagram represents a wave travelling from x to y the wave travels from x to y in a time of 0.5 seconds which row in the table shows the amplitude wavelength and frequency of this wave so first of all we consider the amplitude the amplitude is the displacement of the wave from its rest position vertically to the top or down to the bottom of its crest so here we’ve got a total height of the wave of 2.6 meters which gives us an amplitude from the central line up to the crest or the central line down to the trough of half of that which is 1.3 meters the wavelength while we can see the length there is measured as 12 meters how many waves does that make up well the wave starts here one wave takes us to there two waves three waves four waves so four waves in 12 meters the wavelength is 12 divided by 4 which is 3 meters and finally the frequency well on our data sheet we have a formula that says frequency equals n over t the number of waves divided by time the number of waves as we’ve just said is 4 the time as the question says is 0.5 seconds so 4 divided by 0.5 gives us 8 hertz which gives us an answer of amplitude 1.3 wavelength 3 and frequency 8 which is answer c

10. A microwave signal is transmitted by a radar station.
The signal is reflected from an aeroplane.
The aeroplane is at a height of 30 km directly above the radar station.
The time between the signal being transmitted and the reflected signal being received back
at the radar station is

A 5 × 10−5 s

B 1 × 10−4 s

C 2 × 10−4 s

D 5 × 103 s

E 1 × 104 s.

 

Ans B

Explanation

number 10 a microwave signal is transmitted by a radar station the signal is reflected from an airplane the airplane is at a height of 30 kilometers directly above the radar station the time between the signal being transmitted and reflected signal being received back at the radar station is so there’s our radar station there’s a plane now the radar signal has to travel 30 kilometers to get from the radar station to the plane but then it reflects off the plane so has to travel an additional 30 kilometers back again so if we want to work out the time that it takes we can use distance equal speed times time the distance is 60 kilometers or twice 30 kilometers the speed because these are microwaves part of the electromagnetic spectrum is 3 times 10 to the power of 8. so the time is going to be 60 kilometers or 60 times 10 to the power of 3 meters divided by 3 times 10 to the power of 8 which gives us 2 times 10 to the power of minus 4 seconds answer c

11. A member of the electromagnetic spectrum has a shorter wavelength than visible light and a
lower frequency than X-rays. This type of radiation is

  • A gamma
  • B ultraviolet
  • C infrared
  • D microwaves
  • E radio waves.

Ans : B

Explanation

number 11 a member of the electromagnetic spectrum has a shorter wavelength than visible light and a lower frequency than x-rays this type of radiation is well we need to know the spectrum first of all gamma rays x-rays ultraviolet visible light infrared microwaves tv waves radio waves i remember huge xylophones usually vibrate intensely meanwhile tambourines rattle but what does that tell us well the gamma rays are at the short wavelength high frequency end of the spectrum and the radio waves are the long wavelength low frequency end of the spectrum so we’re looking for the wave that has a wavelength shorter than visible light so we’re looking for the wave that is to the left of visible on the scale there and has a lower frequency than x-rays so lower frequencies towards the right-hand side so the one to the right of x-rays and you’ll see the wave to the right of x-rays and to the left of visible is ultraviolet which is answer b

12. The diagram shows the path of a ray of red light as it passes from air into a glass block.

Ans A

Explanation

number 12 the diagram shows the path of array of red light as it passes from air into a glass block which row in the table shows the angle of instance and angle of refraction well the angle of incidence is the incoming ray so the angle of the incoming ray measured between the light ray and the normal the dashed line there is that one the angle of refraction is the angle between the refracted ray and the normal which is that one so angle of incidence is q angle of refraction is s so the answer is A

13. A sample of tissue is exposed to 15 µGy of alpha radiation and 20 µGy of gamma radiation.
The total equivalent dose received by the tissue is

  • A 35 µSv
  • B 320 µSv
  • C 415 µSv
  • D 700 µSv
  • E 735 µSv.

Ans : B

Explanation

number 13 a sample of tissue is exposed to 15 micrograms of alpha radiation and 20 micrograves of gamma radiation the total equivalent dose received by the tissue is so we’ll consider the alpha first we want to find equivalent dose which is h equals dwr so absorb dose multiplied by the radiation weighting factor the radiation weighting factor for alpha is 20 the absorb dose is 15 micrograys so 15 times 20. i know that radiation weighting factor by the way because it’s given in the data sheet at the front of the exam paper so there you can see alpha is 20. and that gives us an equivalent dose of 300 micro sieverts from alpha we need to do the same with the gamma source though it’s same formula if we look on that table again the radiation weighting factor for gamma is one so we just multiply the 20 micro graves by one which gets us 20 micro sieverts therefore the total equivalent dose is going to be 20 plus 30 which is 3 a 20 plus 300 rather which is 320 micro sieverts answer B

14. Two forces act on an object as shown.

 

The resultant force acting on the object is

A 50 N at a bearing of 053

B 50 N at a bearing of 143

C 50 N at a bearing of 217

D 50 N at a bearing of 233

E 50 N at a bearing of 323.

Ans : C

Explanation

number 14 two forces act on an object as shown the resultant force acting on the object is well if we look at these two forces we can consider that the object will be forced down towards the bottom left of the screen in that sort of direction and we can complete a right angle triangle where the two shorter sides are 40 and 30 and the hypotenuse is the overall force that we’re trying to find out so if we have a right angle triangle we can work out the length of a hypotenuse using pythagoras so the force is going to be the square root of the sum of the squares of the other two sides which is square root of 40 plus square root of 40 squared plus 30 squared which is 50. well the size of the force is 50 newtons but we already knew that because all the options are 50 newtons the direction though well we can work out that angle we have the adjacent and opposite sides of that triangle there tan is opposite over adjacent the opposite is 30 the adjacent 40 so tan of that angle is 30 over 40 or three quarters so the angle x is going to be inverse tan of three quarters which is 37 degrees now that’s not one of my options either but that’s not a bearing that’s why the bearing has to come clockwise from north so if we consider north up there to go from north to south clockwise gets us 180 degrees but then this is a further 37 degrees round from there so 180 plus 37 gives us a bearing of 217. so the correct answer is C 50 newtons at a bearing of 217.

15. The graph shows how the velocity v of an object varies with time t.

The graph could represent the motion of

  • A a ball falling freely downwards
  • B a rocket accelerating upwards
  • C a ball thrown into the air then falling back to Earth
  • D a ball falling to Earth from rest then rebounding upwards again
  • E a car slowing to a halt then accelerating in the same direction.

Ans C

Explanation

number 15 the graph shows how the velocity v of an object varies with time the graph could represent the motion of well what’s happening here initially we’ve got the velocity decreasing from some value towards zero and it’s a positive value it’s above zero then at that point there the velocity is zero so the object would instantaneously stop and then you can see on the other side the line on the graph goes from positive to negative that indicates a change in direction and here we have the line moving away from zero so increasing but in the negative direction so something is slowing down in one direction until it stops and then speeds up in the other direction so which of the options would satisfy that criteria well a ball falling freely downwards would not change direction it continues to fall in one direction so it’s not that a rocket accelerating upwards accelerates in one direction upwards a ball thrown into the air then falling back to earth well if we imagine throwing a ball up into the air it would slow down as it gets higher until it gets to the highest point of its trajectory and then it would start to speed back up again on the way down so answer c is correct b a ball falling to earth from rest and then rebounding upwards again well if it fell from rest then the line would be sloping upwards as it increases its speed towards earth so it’s not b and the car’s slowing to hop then accelerating in the same direction well it doesn’t have a change in direction it would slow down and then speed up again but the line would never cross that horizontal axis of the graph the answer is three

16. A trolley is released from rest at point X and moves with constant acceleration on a slope as
shown.

The computer displays the acceleration and average velocity of the trolley between the light
gates.
The trolley is now released from rest at point Y.
Which row in the table shows how the acceleration and average velocity compare with the
previous results obtained?

Ans :E

Explanation

number 16 a troll is released from rest at point x and moves with a constant acceleration on a slope as shown the computer displays the acceleration and average velocity of the trolley between the light gates the trolley is now released from point y which you can see is further down the slope which row in the table shows how the acceleration and average velocity compare with the previous results obtained well if it’s accelerating from y rather than x the slope itself doesn’t change so therefore the force that’s pulling the trolley down the slope is not going to change and therefore there will be no change in the acceleration of the trolley however because it has been accelerating for less time when it reaches the like gates it will not have been going as fast by the time it gets to the light gate and therefore it will be moving with a lesser average velocity so the answer is e

17. A rocket accelerates vertically upwards from the surface of the Earth.
An identical rocket accelerates vertically upwards from the surface of Mars.
The engine thrust from each rocket is the same.
Which row in the table shows how the weight of the rocket and the unbalanced force acting
on the rocket compares on Mars and Earth?

Ans : D

Explanation

number 17 a rocket accelerates vertically upwards from the surface of earth an identical rocket accelerates vertically upwards from the surface of mars the engine thrust from each rocket is the same which row in the table shows how the weight of the rocket and the unbalanced force on the rocket compares on mars nerf so what is the difference between mars and earth if we look on our data sheet we have this table which tells us the gravitational field strength on different planets and we see earth has a gravitational field strength of 9.8 whereas mars has a gravitational field strength of 3.7 newtons per kilogram that tells us that the force of gravity acting on the rocket on mars is a little bit less than half that on earth so let’s consider the weight is going to be less on mars because the gravitational field strength and less what about the unbalanced force well if we think about earth and mars and we’ve got this identical rocket with an identical thrust on earth the weight is large on mars this weight is less a smaller weight means that the unbalanced force so the difference between the thrust and the weight on mars is going to be greater than the difference between the thrust and the weight on earth because they have the same thrust but the earth’s weight is larger than mars’s weight answer d

18. A satellite is in a circular orbit around a planet.

A group of students make the following statements about the satellite.
I The greater the altitude of a satellite the shorter its orbital period.
II The satellite has a constant vertical acceleration.
III As the satellite orbits the planet, its vertical velocity increases.
Which of these statements is/are correct?

  • A I only
  • B II only
  • C III only
  • D I and II only
  • E II and III only

Ans B

Explanation

number 18 a satellite is in a circular orbit around the a planet a group of students makes the following statements about the plan about the satellite the greater the altitude of a satellite the shorter its orbital period the satellite has constant vertical acceleration and as the satellite orbits the planet its vertical velocity increases which of these statements is or are correct well firstly the greater the altitude of the satellite the shorter its orbital period is not correct the further out the satellite is so the greater its altitude the longer its orbital period would be the satellite has a constant vertical acceleration that is true the satellite is in free fall around about the earth it is being influenced by the earth’s gravitational force um is causing it to accelerate towards earth at the gravitational acceleration at the height that it is at as the satellite orbits the planet its gravity its vertical velocity increases that’s not true although it has a vertical acceleration that does not result in an increase in its vertical velocity what happens is remember velocity is a vector it has both magnitude and direction an acceleration is a change in velocity over a period of time that change in velocity does not necessarily have to be in the size of the velocity it can be in the direction of the velocity and that’s what happens in the satellite so although it is accelerating towards the earth that acceleration does not alter the magnitude of the satellite’s velocity it alters the direction of the satellite’s velocity and therefore causes it to orbit the planet so that is not correct the only one that is correct is two so answer b

19. A heater transfers energy to boiling water at the rate of 1130 joules every second.
The maximum mass of water converted to steam in 2 minutes is

  • A 1·0 × 10−3 kg
  • B 6·0 × 10−2 kg
  • C 0·41 kg
  • D 17 kg
  • E 32 kg.

Ans B

Explanation

number 19 a heater transfers energy to boiling water at a rate of 1130 joules every second the maximum mass of water converted to steam in two minutes is well we’re changing the state of the water so we’re going to use our formula eh equals ml where l is the latent heat of vaporization of water which we can get on our data sheet and we can see it is 22.6 times 10 to the power of 5 joules per kilogram so we can insert these numbers into our formula and we get a mass of 5 x 10 to the power of minus 4 kilograms that however is the mass that is being converted to steam each second because it tells us that the energy that’s being transferred to the water is 1130 joules each second so if that amount of mass is converted each second and the question asks us how much mass is converted in two minutes we need to multiply that by the number of seconds in two minutes which is 120 which gives us a total mass of 0.06 kilograms which is answer b

20. Light from stars can be split into line spectra of different colours.
The line spectra from three stars, X, Y and Z, are shown, along with the line spectra of the
elements helium and hydrogen.

Hydrogen and helium are both present in

  • A star X only
  • B star Y only
  • C stars X and Y only
  • D stars X and Z only
  • E stars X, Y and Z.

Ans : D

Explanation

the final question in this paper light from stars can be split into line spectra of different colors the line spectra from three stars x y and z are shown along with the line spectra of the elements helium and hydrogen hydrogen or helium are both present in which stars well how are we going to know we will know if the lines present in helium and hydrogen are all present in a star then that element is present in the star so let’s consider helium first of all and look at the line spectra working from left to right so looking at the leftmost line of helium we see that that line is present in star x y and z if we look at the next line in helium we’ll see that line is also present in star x y and z indeed so is the next line and the next line and the next line so helium is present in star x star y and stars red now let’s consider hydrogen there are only two lines in hydrogen so let’s contin consider the first one first so that line there is present in star x and star y that’s fine start x on star z but not in star y because it’s not in star y if one of these lines is not present in star y then that means hydrogen is not there so star y contains helium only let’s consider the other line of hydrogen to see if it’s present in star x or z so the other line of hydrogen we can see lines up with one of the lines in star x and one of the lines in star z so both star x and star z also contain hydrogen star zed you can see has another set of lines as well so star x contains only hydrogen and helium because hydrogen and helium account for all the lines in star x star zed contains hydrogen and helium and some other element which accounts for the other three lines not already accounted for so the answer is star x and star zed which is answer D

 

and that completes the SQA National 5 Physics 2017 section one i hope you found this helpful and we will have a look at section two of the paper and the different topics on Website.

 

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