In this article, I will be discussing a series of questions from the 2017 Henry Park Primary School (HPPS) P5 SA2 Science Examination Paper. I will be sharing my analysis behind each P5 Science topical question so that your child can have a better idea of how to tackle similar open-ended questions in the future.

One of the biggest challenges that I have noticed for many Primary 5 students especially is that when they move up from Primary 4 to Primary 5, the standard of questioning and their expected answers rises quite significantly.

It is not so much the concepts that they struggle with, but the application of the concepts that they have learned.

I will be able to show and share more with you as I work through these P5 Science topical questions from the following topics:

So, without further ado, let us get started!

Q29 (Heat Energy)

Q30 (Heat Energy)

Q31 (Man’s Impact on the Environment)

Q32 (Electricity)

Q33 (Electricity)

Q34 (Reproduction)

Q35 (Reproduction)

Q36 (Transport in Plants)

Q37 (Matter)

Q38 (Heat Energy)

Q39 (Light Energy)

Q40 (Animal Life Cycle)

Q41 (Magnets)

## Let’s Take A Look At This Question – Q29 [Heat Energy]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q29]

This question is about the factors of evaporation.

What are the four factors of evaporation and how do we remember them?

I teach my students to follow the acronym WHAT.

W stands for Wind.

*Do note that teachers will not accept just “Wind” as a word on its own. They want the students to write “presence of wind.” When it comes to comparison terms we must also state “stronger presence of wind.” Or “weaker presence of wind”.

H stands for Humidity. Humidity is not in our syllabus so it is unlikely that we will have to use it in the answer.

A is Area of exposed surface or some students may call it Exposed surface area.

T stands for Temperature. The temperature’s source must be specified. Is it the temperature of the substance itself? Or is it the temperature of the surrounding air?

You might be wondering what the difference is between temperature of the substance and temperature of the surrounding air.

If I have a cup of hot water and I place it on a table, I cannot say that “the hot water gains heat from the surrounding air” because the hot water is hotter than the surrounding air. However, if the surrounding air is warmer than the water, I can say that the cup of water gained heat from the warmer surrounding air to evaporate.

That is the difference; In one situation I need to state that heat was gained, while the other situation does not require any statement of heat gained. I just say the hot water in the cup will evaporate to form water vapour.

### Let Us Take A Look At The First Question – Q29a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q29a]

Which of the four factors of evaporation are we going to focus on?

One is folded, one is not, therefore this question is about exposed surface area.

Why did T-shirt R dry faster?

Because T-shirt R has a larger exposed surface area. The water in the T- shirt “gained heat faster from the warmer surroundings” or you can say “gained heat faster from the sun to evaporate faster”.

This is the suggested answer for part (a):

A larger exposed surface area of water is in contact with the surrounding air. Water in the T-shirt to gain heat faster from the warmer surrounding air to evaporate faster.

One key thing that I want you to take note from the answers is this, “the water in the T-shirt” because some teachers are particular about the phrasing.

Even if you say that the T-shirt evaporates, they would not give you the full marks because it’s the water in the t-shirt that evaporates.

### Let Us Look At The Next Question – Q29b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q29b]

If you look at the hanger in Figure 2, this hanger with the figure 8 actually creates more air space within the t-shirt itself as compared to Figure 1.

This question again tests the concept of exposed surface area. Therefore our answer to part (b) is:

Because of the hanger in figure 2, there is a larger exposed surface area of the water in the t-shirt in Figure 2 that is in contact with surrounding air. The water in the t-shirt will gain heat faster and then evaporate faster.

There is an alternative answer: It can also be argued that “because the T-shirt in figure 2 has a larger exposed surface area, it is more “airy”. More wind is able to circulate inside the T-shirt. Presence of wind in the T-shirt will cause water to evaporate faster.

### Now, Moving On To The Last Part – Q29c:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q29c]

Let us recall. For an experiment to be a fair test there can only be one changed variable.

What was the changed variable in this question?

The changed variable is the exposed surface area of the T- shirt. How does conducting the experiment in the same place ensure that it is a fair test?

Recall the four factors of evaporation: Presence of Wind, Humidity, Area of exposed surface, and Temperature.

If we are going to change the exposed surface area of the T-shirt, the rest of the factors that affect the rate of evaporation should be kept the same. For our answer for part (c), we are going to talk about temperature and presence of wind.

The purpose of the experiment is to see how the method of hanging the T-shirts affects the time taken for the T-shirts to dry. By conducting each experiment in the same location, we ensure that the temperature of the surrounding air and the presence of wind, or the wind speed, is kept the same and thus will not affect the results.

Note that we must also remember to specify what variable we are changing and what results are being recorded.

## Let’s Take A Look At This Question – Q30 [Heat Energy]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q30]

### Let’s Tackle The First Part – Q30a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q30a]

The diagrams tells me that this question aims to test students about hot water and cold water.

This is where we use a bit of common sense. If you have a cup of water, which is cold, and you leave it out in the room at room temperature, what will you notice about the cup?

I would start to observe water droplets forming on the outer surface of the cup. If we had a cup with hot water instead, where would the water droplets be formed? The water droplets will be formed on the inner surface of the cup.

With that in mind, I know that my answer for (a) will be:

P represents cold water and Q represents hot water.

### Now, Moving On To The Next Part – Q30b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q30b]

The key phrase here is “to increase”.

Picture this scenario: I have two cups of water. Both cups of water are cold and I leave it on the table on a room at room temperature. How would I know which cup is colder? What would I observe? What is the difference?

I would observe that the cup with more water droplets formed on the outside would be colder.

Therefore, how do we increase the number of water droplets in set-up P?

With the earlier scenario in mind, we decreased the temperature of the water in set-up P’s beaker so that it becomes colder.

Next, how do we explain how the water droplets form on the outer surface of the beaker in the first place?

The warmer water vapour from the surrounding air comes into contact with the cooler outer surface of the beaker, loses heat to the cold surface and condenses to form tiny water droplets.

Don’t forget that we must also state how we make the beaker colder!

First, you add ice to the water in the cup so that the water in the set-up P will lose heat to the ice and decrease in temperature. Once the water temperature has decreased, the beaker will then lose heat to the cooler water so the beaker now becomes even colder. When the warmer water vapour in the surrounding air comes into contact with the beaker’s cooler outer surface, it loses heat faster and quickly condenses to form more water droplets.

The key word here is the comparison term FASTER.

## Let’s Take A Look At This Question – Q31 [Man’s Impact on the Environment]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q31]

Question 31 may be quite confusing for some P5s because it is a “Man and the Environment” question. This is a topic that will only be taught in P6! However, Henry Park has decided to bring this topic down to the P5 level, so I will try my best to explain the context of the question in a simple way.

Factories may sometimes dump industrial and machine waste into the river. At times, they may also release poisonous gases through their chimneys.

When farms use excessive amount of fertilisers or insecticides, some of these chemicals will seep into the soil and when it rains, the water mixes with the chemicals in the soil and gets washed into the river causing pollution as well.

With that context in mind, let us proceed to the question:

### Let’s Tackle The First Part – Q31a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q31a]

We must take note of the direction of the river flow. It is flowing downstream. If you look at P, it has yet to pass by the farm and the factory. Q, is where the farm and the factory are located and then R is the water after the farm and the factory.

Let us look more closely at the information given.

If you are looking for the least polluted sample, you should look at the water sample where the fish were unaffected. This is because it is likely that the fishes died due to the pollution that came from the farm and the factories.

Based on this information given, I will choose Point P because none of the fishes died.

To answer the question, follow our C-U-E method of writing. C stands for Choose. U stands for Use the data – which is the evidence in the question. And E is the explanation.

This is how we would phrase our answer to part (a):

I’m going to Choose “Point P”. Using the data means using whatever that they have given us in the question. “None of the fish died at Point P”.

Now, we will attempt to explain: “This shows that the water has not been polluted by the factory and the farm and there were no harmful substances in the water that affected the fishes. And therefore, they survived.

### Let’s Move On To The Next Part – Q31b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q31b]

In order for us to understand global warming, we must understand what causes global warming. One of the key causes is actually greenhouse gases and one major greenhouse gas that we always talk about is carbon dioxide.

How is this linked to trees?

First of all, when trees are cut down, there will be less trees to take in carbon dioxide from the surrounding air for photosynthesis to make food. Which means there will be more carbon dioxide left in the surrounding air. They will accumulate. That’s how carbon dioxide increases.

The other factor is the action of burning the trees. This causes a lot of carbon dioxide to be released into the environment. When there is more carbon dioxide, which is a greenhouse gas, then more heat from the sun will be trapped by these greenhouse gases, and this will lead to Global Warming.

Either factor is a viable answer to this question. Here is how I would phrase my answer to part (b):

There will be less trees to take in carbon dioxide for photosynthesis to make food. carbon dioxide is a greenhouse gas, and when it accumulates, it will trap more heat from the sun. This leads to global warming.

Remember that you MUST state that carbon dioxide is a greenhouse gas!

## Let’s Take A Look At This Question – Q32 [Electricity]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q32]

Questions on electricity often feature circuit diagrams arranged in series or parallel configurations. In order for us visualise the results shown in the graph, let me draw an example diagram for a parallel circuit first.

### Let’s Tackle The First Part – Q32a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q32a]

In this diagram, we have two batteries and three bulbs. How do we count the brightness of the bulbs?

I always encourage my students to trace the diagram when working with circuits.

In this diagram below, I have outlined one circuit in red:

How many batteries do I have? Two batteries.

How many bulbs do I have? One bulb.

Two batteries, one bulb means that the power given to this bulb here is 2.

Let me draw the next mini circuit that I have. If you follow the blue line:

How many batteries do I have here? Two batteries.

How many bulbs do I have? One bulb.

Therefore, the power is again 2.

The last circuit is traced in green this time.

Again, I have two batteries and one bulb.

The power is also 2.

If the power given to each of the bulbs is the same, then the brightness of the bulbs is going to remain constant. Based on that observation, we can tell that Graph X represents bulbs in parallel.

In the second graph, the brightness of the bulbs slowly decreases. Let me give you a scenario.

In this red circuit, we have two batteries, one bulb. What is the power given to it?

If you look at the previous set-up, using the same style of analysing the question, we can say that the power of 2 is given to this bulb here.

But what happens now if we were to give two batteries and two bulbs in series?

In this circuit here, we have two batteries, two bulbs. Do they need to share?

Yes, they need to share. Two divided by two, each one will get a power of 1. With less power given to each bulb, the brightness will decrease.

Let us now add a third bulb to the circuit.

Now we have two batteries and three bulbs. Each bulb will now receive a power of 2/3. It will be even dimmer than when there were two bulbs.

As you add more and more bulbs in series, what you notice is that the electricity has to be shared by each of the bulbs added. And the brightness of the bulbs will start to decrease.

Therefore, Graph Y on the right side, will likely to represent the series circuit.

Returning to the question: “Based on the information given, state how the bulbs are arranged, series or parallel.”

X is parallel circuit and Y is the series circuit.

### Let’s Tackle The First Part – Q32b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q32b]

Is it accurate to use our eyes to measure the brightness?

Perhaps not. While we might be able to observe that one bulb is brighter than the other, we will not be able to tell quantifiably how much brighter because brightness is really subjective to how your eyes receive the light.

A good way to determine how much light is produced by each bulb is to use a light sensor.

Therefore the answer to part (b) will be:

Use a light sensor with a data logger to receive the light. A light sensor will be able to measure the amount of light emitted by each of the bulbs accurately.

## Let’s Take A Look At This Question – Q33 [Electricity]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33]

As before in Question 32, we should start by tracing the electrical path of each circuit.

Let us start up with Circuit A:

As illustrated above, there is one possible pathway for electric current to pass through Circuit A.

Let us look at Circuit B. How many possible pathways are there for electric current to flow through the circuit?

There are two possible pathways. One is on the inside. And one is on the outside. I have used different colours to represent them for clarity. The one on the inside is the purple line. The one on the outside is the red line.

Lastly, Circuit C.

There is one possible pathway for electric current to flow through.

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33]

### With That, Let Us Move On To Q33a

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33a]

Now I want you to think about this analogy: I have five sweets and I want to distribute them to two people.

Firstly, how many people are there altogether? Two.

Therefore, if I were to distribute five sweets to them equally, each one will get two and a half sweets. How do I get two and a half? Five divided by two.

Using the same idea and the same concept, let us look at Circuit A again.

How many batteries do I have? Three batteries.

How many bulbs do I have? Two bulbs.

All the electric current is going to come from the three batteries. Each bulb would then receive electric current from how many batteries? Three divided by two gives us one point five (1.5). I will call this a power of 1.5 received from the three batteries.

Let us look at Circuit B

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33] – Circuit B

Let us start off with the purple circuit:

How many batteries do I have in the purple circuit? Three batteries.

How many bulbs do I have? Only one bulb.

What does this tell us about Circuit B’s purple circuit?

It means that it does not need to share. Bulb R is going to receive all the electric current comes from the three batteries. The power given to R is 3.

Let us look at the one below it. Bulb S. We trace the red line here and we notice that there are three batteries in the red line. How many bulbs are there? One bulb as well. It does not need to share. The power received by S is also going to be 3.

The last one for C.

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33] – Circuit C

This is slightly different. We have three batteries. How many bulbs do we have? We’re going to have three bulbs. Three batteries, three bulbs. Do they need to share? The answer is Yes. Three divided by three, each one will get 1.

Returning to the question: “Based on the information given, arrange bulbs P, R and T in order of brightness, from the brightest to the dimmest.”

The brightest one is going to be the one that has the largest power number, because it receives the most electric current. It will glow the most brightly.

Therefore, based on our observations above, the answer to part (a) will be in order:

R, P, and T

### Moving On To The Next Part Of The Question – Q33b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33b]

The advantages and disadvantages of series and parallel circuits is a common topic in questions about electricity.

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33] – Circuit B

Circuit B has bulbs arranged in parallel. If bulb R fuses or is removed, one circuit (outlined in purple) is broken, but there is still a closed circuit with bulb S (outlined in red). This means the electric current can still flow through bulb S. This allows the bulb S to be lit.

This is our answer for part (b):

Bulb S will remain lit.

### Lastly, Here’s How To Tackle – Q33c:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q33c]

I am going to introduce to you the four steps to answering such a question:

Step 1) What is the situation?

Step 2) Is this now an open or closed circuit?

Step 3) Does electric current pass or flow through?

Step 4) What is the outcome?

Let me show you the answer to part c and I will discuss it from there.

The bulbs in circuit B are arranged in parallel. Therefore even when bulb R is removed, there is still a closed circuit with bulb S, allowing it to remain lit.

Step 1 is the situation: “When bulb R is removed”.

Step 2 is the question of whether the circuit is now open or closed. (yes)

Since the answer to Step 3 is yes, we now look at whether bulb S lies within the closed circuit where electric current is passing through. (yes)

Therefore, the outcome is “Bulb S will remain lit”.

## Let’s Take A Look At This Question – Q34 [Reproduction]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q34]

Since the question talks about the function, let us quickly recall some facts:

What are stomata and where do we find them?

Stomata are tiny openings found mostly on the underside of the leaf.

What is the key function of the stomata?

They allow for the exchange of gases.

The first one is oxygen. Oxygen is taken in through the stomata for the plant to carry out respiration to release energy and it releases carbon dioxide through the stomata as well.

The second gas is carbon dioxide. carbon dioxide is taken in by the plant for photosynthesis to make food.

The last gas is water vapour. Water vapour is lost through the stomata in the process of Transpiration.

These three gases, oxygen, carbon dioxide, and water vapour, are involved in three processes – photosynthesis, respiration and transpiration.

Next, let us discuss the gill filament.

The gill filament in fish is similar to the lungs in humans and animals.

The lungs allow for the exchange of gases. The gills allow for the exchange of gases as well. You will observe many lines in the diagram. These are gill filaments that are functionally similar to the air sac in our own lungs.

Air sacs increase the exposed surface area of the lungs for faster exchange of gases. This is mainly oxygen and carbon dioxide. Similarly, the gill filament increases the exposed surface area of the gills for faster exchange of dissolved gases.

Note that we mention dissolved gases because the fish takes in dissolved gases from the water. It is not able to take in gases from the surrounding air.

### Let Us Tackle Q34a Together

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q34a]

Returning to the question: “State one similarity in the functions of both the stomata and the gill filament.”

As per the above explanation the answer for part (a) is:

Both allow for gaseous exchange to take place.

### Now, Let’s Move On To Q34b

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q34b]

As mentioned earlier, having numerous gill filaments is like having numerous air sacs. The key point is the exposed surface area.

The gill filaments increase the exposed surface area of the gills in contact with the water so that it can have faster exchange of dissolved gases.

## Let’s Take A Look At This Question – Q35 [Reproduction]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q35]

Question 35 is rather straightforward.

### Here’s How We Can Answer Q35a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q35a]

R represents the “Ovary”. If they have labelled two sides, then some of your school teachers might want you to specify and call it “Ovaries.

Q represents the “womb”.

### Moving On To Q35b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q35b]

Part R is the ovary. Putting it simply, for part (b) of our answer:

Part R produces and releases the egg cell.

This is just a little bit more information: The egg (sometimes rarely ‘eggs’) go into the fallopian tube. At the fallopian tube, the egg fuses with the sperm in the process of fertilisation. When the fertilised egg is formed, it will move down to the womb where it is implanted and develops into a baby.

### Lastly, Let’s Take A Look At Q35c:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q35c]

I want you to recall some of the concepts that we learned under the topic of cells. Let us specify the functions of this particular cell part, which is the nucleus.

The nucleus controls all the activities in the cell, that is the first one. It also contains genetic information which is passed down from one generation to the next.

When we talk about passing down from one generation to the next, we recall that the egg comes from the mummy, Mrs Chan. The sperm comes from daddy, which is Mr. Chan. For a Primary 5 student, it is sufficient to know that the genetic information is found in both the egg and the sperm and they fuse together. This is why Billy will receive genetic information from both his parents.

Genetic information from both his parents were passed down to Billy, allowing him to inherit characteristics from both his parents.

## Let’s Take A Look At This Question – Q36 [Transport In Plants]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q36]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q36a]

Note that this statement is very similar to questions that ask to specify the aim of the experiment. The keywords are “How something affects something.

Whenever we see something like that, we will put a box around the keyword “how”, and the keyword “affect” and then we’re going to underline anything in between the word “How” and the word “Affect” and then anything from the word “Affect” to the full stop. This is important for us to identify the two variables.

In any experiment, there are two variables. One is called the changed variable. And the other is the measured variable.

The changed variable is whatever we are testing. The measured variable is essentially our results.

In this question, the size of the flowers is the changed variable and the number of butterflies landing on the flowers is the measured variable.

“John placed the four flowers in the garden and counted the number of butterflies that landed on each flower.”

Why did they specify the garden?

Because this is to ensure that the location was kept the same so that there is only one changed variable. This helps to ensure a fair test.

Based on the results provided in the table, we noticed that the flower with size J (which is the largest) has the most number of butterflies landing on it. Meanwhile, the size M flower (which is the smallest) has the least number of butterflies landing on the flower.

This is how we will phrase our answer.

### Next, Let’s Move On To Q36b

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q36b]

IF is my keyword. “If” something affects something.

Now, we have the two variables again.

The measured variable, the number of butterflies visiting the flowers, is the same as in part (a) and the table that they have given you. But now, the testing (change) variable is different. Instead of the size of the flower, we are now testing the colour of the flowers.

Because we are no longer testing the size of the petals, the new petals must all be the same size. But, we must also make sure that the colour of the flowers are different.

Change 1: Make the size of the petals for all the flowers the same.

Change 2: Change the colour of the flowers for each plant to a different colour.

### Lastly, Let’s Have A Look At Q36c:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q36c]

Butterflies are known as pollinators. Pollinators are animals or insects that visit the flower and when they visit the flower, usually it’s to obtain nectar. Their bodies will rub against the anthers. The anthers are the parts that produce and store the pollen grains. The pollen grains from the anther would then stick on to the insects’ body. And when the insect visits another flower of the same species, the pollen grains would then land on the stigma in the process of pollination.

Once the flower is pollinated, fertilisation will take place and flower develops into a fruit. With the fruit, then there will be seeds. And with seeds, then the seeds will then germinate and form a new plant and this is how the plant ensures the continuity of its own kind.

Butterflies aid in the process called pollination. When a butterfly visits a flower for nectar, they also collect pollen when their bodies rub against the flower’s anthers. They then carry those pollen grains to the next flower where the pollen is deposited on the stigma for fertilisation. The fertilised flower develops into a fruit, which in turn will have seeds that can germinate and form new plants of the same species.

## Let’s Take A Look At This Question – Q37 [Matter]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q37]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q37a]

This is a question on the topic of matter.

Matter has mass and it occupies space. When you place the pebble to the water, the water will get displaced, which means that the pebble occupies space and the water, the water has to move upwards. The reading of the water level increases but the amount of water remains the same.

This is how we’re going to phrase our answer for part (a):

No. Liquids have a definite volume and therefore, the volume does not change. The volume of the water would remain the same, but the water level rises because the pebble also occupies space in the cylinder. This forces the water to move upwards.

For every pebble added, the water level increases by 20 units. So, we can then conclude for that one pebble has a volume of 20 units.

### Move On To The Next Part – Q37b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q37b]

First of all, let us try to understand what the question is all about. When the water is flushed, the water level decreases and then the water starts to fill up the tank to level L.

As per part (a), if we were to put some pebbles into the water, the pebbles would occupy the space of the water which will cause the water level to increase. However, the volume of water in the cylinder, or in this case, the tank, remains the same.

If we were to put more pebbles into the water tank, the amount of water needed to fill the water tank to water level L, would be less.

This is how we will phrase our answer:

The pebbles would occupy space in the water tank. What happens now is that   less water would be needed to fill the water level in water tank to reach level L, and that’s how Beng Soon is going to conserve the water.

## Let’s Take A Look At This Question – Q38 [Heat Energy]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q38]

### Let Us Take A Look At Q38a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q38a]

In this question, we are given three materials to consider: Paper, foil, and bubble wrap. We also have one heat source: The hot milo.

We have learned that different materials have different heat conductivities. A material which is a better conductor of heat would allow heat to pass through it faster as compared to a poorer conductor of heat where heat to passes through slower.

Let us take a closer look at the table:

Notice that the lowest temperature (27°C) was reached after 20 minutes using the paper. This means that when the cup is wrapped with paper, the heat from the hot milo was lost to the cooler surroundings at the fastest rate.

Note that I used the word “fastest” instead of “faster” because this question compares three materials. If we were just comparing with two materials, I would use “faster” rate.

On the opposite end of the list, the milo in the bubble wrapped cup was at 50°C after 20 minutes. That is the highest temperature reading after 20 minutes. This means that heat from the hot milo and the bubble wrapped cup was lost to the cooler surroundings at the slowest rate.

Returning to the question: “Based on her results, which material (paper, foil or bubble wrap) will keep her Milo hot for the longest period of time?”

With any choosing question, we must remember to use our CUE.

C stands for Choose.

U means Use the Data – the evidence in the question. In this case, I will talk about temperature of the milo after 20 minutes.

stands for Explanation, which means that we must link the data to the Science concept.

Let me show you the answer for part (a) while breaking it down.

I choosethe bubble wrap.”

When using the data (from the table), there are two ways you can phrase it. One way is to state that “the temperature of the milo after 20 minutes, it was the highest” or “the decrease in the temperature of the milo and the cup with the bubble wrap was the smallest.” If the decrease is the smallest, it means that the least amount of heat was lost.

Now we’re going to explain using the science concept. “This shows that the bubble wrap is the poorest conductor of heat. Thus, it would allow heat from the milo to be lost to the cooler surrounding air the slowest and keep the hot milo hot for the longest period of time.”

### Now, Moving On To The Next Part – Q38b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q38b]

When we look at heat lost to the cooler surroundings, we expect that the temperature decreases. However, in the table above, the temperature drops from 63°C to 25°C and then rises to 45°C.

While it is possible for 63°C to decrease to 25°C, there is no heat source or constant flame to provide more heat, it is therefore impossible for the milo to warm back up to 45°C.

The temperature is 25°C because the temperature of the water cannot decrease and then increase to 45°C again.

## Let’s Take A Look At This Question – Q39 [Light Energy]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q39]

Fact: The closer the light is to the object, the larger the shadow is cast.

### Let Us Take A Look At Q39a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q39a]

Move the light source nearer to the puppets.

Generally, think about how shadows are formed. There is a template structure for shadows. We say that “Shadows are formed when the path of light from the light source, which travels in a straight line is blocked by an object, which is opaque or translucent.” Opaque and translucent objects block light. As long as light is blocked, shadows can be formed.

### Now, Let’s Move To Q39b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q39b]

In this case the answer for part (b):

The shadows of the puppets are formed when light from the light source, which travels in a straight line, is blocked by the puppets, which are opaque.

### Let’s Take A Look At Q39c:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q39c]

A material that allows most light is a transparent material. An example of a transparent material is glass. Shadows cannot be formed on glass, therefore the answer is no, she is not correct.

The reason: “A material that allows most light to pass through would not allow shadows to be formed on it.

This is unlike an opaque material (allows little to no light to pass through) and a translucent material (allows only some light to pass through).

This question revisits the P4 topic of animal life cycles. A mosquito has a 4-stage life cycle: Eggs -> Larva -> Pupa-> Adult.

## Let’s Take A Look At This Question – Q40 [Animal Life Cycle]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q40]

In this case, the answer is:

Part (i):“pupa”

Part (ii) “larva”.

### Next, We’ll Move On To Q40b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q40b]

Whenever we answer a relationship question, we must put a box around the key words “between” and “end”. This gives us our two variables. The first one is temperature. The second one is the average length of the life cycle of an Aedes mosquito.

Looking at the table provided, as the temperature increases, the average length of the life cycle of an Aedes mosquito goes down. We say it decreases.

As the temperature increases, the average length of the lifecycle of an Aedes mosquito decreases.

### Lastly, Let’s Have A Look At Q40c

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q40c]

Based on answer for part (b), we know that when the temperature is higher the average life cycle of the Aedes mosquito is shorter. This means that the eggs hatch more quickly and the larva proceeds through the stages of pupa to adult faster. In other words, it becomes an adult at a faster rate. Which also means the rate of the reproduction is also faster. There would be likely to have more mosquitoes in the environment.

During the warmer months, the life cycle of the mosquito is shorter, so they reproduce faster and thus, there would be a larger number of adult mosquitos which would spread the dengue virus at a faster rate.

## Now, We’re Down To Our Final Question – Q41 [Magnets]

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q41]

### Let Us Get Started With Q41a:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q41a]

This last question seems tricky, featuring different lengths of magnet and different amounts of force required to pull the magnets apart. It is actually straightforward:

What is the smallest force needed to pull the bar magnets apart?

This means that smaller amount of force needed, the weaker the bar magnet.

The answer for 41(a) is therefore:

Magnet “B”.

### Moving On To The Next Part, Q41b:

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q41b]

Let us consider how magnets are created. There are two methods:

1) Stroking a magnetic material with another magnet. As we increase the number of times we stroke the material in one direction, the magnetic strength or the bar magnet becomes stronger.

2) Construct an electromagnet with a circuit. By increasing the number of batteries used in the circuit or winding more coils around the magnetic material, the magnetic strength also increases.

With those two principles in mind, we know that the length or the size of the bar magnet is not going to affect the strength of the magnet.

The question, however, requires that we make use of the data provided. Let us take another look at the table.

There are two ways that we can go about answering this question. We can either compare bar magnet A and B. Or we can compare C and D.

Let us compare A and B. Based on Caili’s hypothesis, the shorter magnet, A, should be weaker. However, if you look at the amount of force needed to pull the magnets apart, A actually required a greater amount of force. This means that bar magnet A is shorter but its magnetic strength is also stronger.

Likewise, if we compare C and D. C has a shorter length than D. Based on Caili’s hypothesis, you would expect that bar magnet C’s magnetic strength to be weaker than that of D. But the force needed to pull magnet C apart from the horseshoe magnet was 8, as compared to D’s 4. This means that the magnetic strength of bar magnet C is stronger than that of D.

For my suggested answer for part (b), I compared C vs D.

Although Bar magnet D was longer than bar magnet C, the amount of force needed to pull bar magnet D apart from the horse shoe magnet was smaller. This shows that the longer bar magnet D’s magnetism was weaker than the shorter Bar Magnet C’s magnetism.

Make sure that you use the comparison terms. Do not just provide the values. They are not conclusive without the comparisons.

### Last But Not Least, Let’s Tackle Q41c

Source: Henry Park Primary School – 2017 P5 SA2 Science Examination Paper [Q41c]

This is a very simple experiment. Some children might even have done this in some of their school’s science practical.

Magnetic strength can also be determined by the maximum distance from which the bar magnet can attract the paper clip. This is because a magnet’s magnetism can act at a distance. The stronger the magnet, the further it can act or attract other magnetic materials.

This is my answer for part (c):

Using the ruler and Caili can measure maximum distance from which the bar magnet is able to attract the steel paper clip. A bar magnet with a greater magnetic strength would be able to attract the steel paper clip from a further distance.

And that concludes the answers and the discussion for the 2017 Henry Park Primary Paper P5 SA2 Science Examination Paper.

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