The Oxford PAT (Physics Admissions Test) is around the corner. And as is always the case with physics, the most effective practice method involves solving problems. This blog also offers numerous problem-solving exercises to aid in enhancing your skills.
While we can’t guarantee success in the PAT just by practising these problems – your score ultimately relies on your efforts – thoroughly working through all the problems and understanding their solutions will also significantly strengthen your preparation for the PAT when the time comes.
Oxford PAT Questions: #1
A jar contains buttons of four different colours. There are twice as many yellow as green, twice as many red as yellow, and twice as many blue as red. What is the probability of taking from the jar:
- a blue button;
- red button;
- yellow button;
- green button?
You may assume that you are only taking one button at a time and replacing it in the jar before selecting the next one.
To get the most out of this question, start by giving it a good read, including the hint. Furthermore, take your time to think about your answer. After you’ve pondered it, feel free to check out the solution below.
Hint:
Assign variables to the numbers of each colour of button and try to convert the information in the question into a set of equations.
SOLUTION #1:
Let nb, nr, ny and ng be the numbers of blue, red, yellow and green buttons respectively.
Also let N = nb + nr + ny + ng be the total number of buttons.
Firstly, we need to convert the information given in the question into mathematical statements. The fact that there are “twice as many yellow as green” means that ny is twice ng, or equivalently ny = 2ng.
Similarly, “twice as many red as yellow” becomes nr = 2ny
And “twice as many blue as red” becomes nb = 2nr.
This therefore gives us:
ny = 2ng (1.1)
nr = 2ny (1.2)
nb = 2nr. (1.3)
Although we don’t know the total number of buttons N, equations 1.1–1.3 allow us to figure out the ratios of buttons. For example, if there were one green button, then by equation 1.1 there would be two yellow buttons.
If there were two yellow buttons, then by equation 1.2 there would be four red buttons. However, if there were four red buttons, then by equation 1.3 there would be eight blue buttons. The ratio of green : yellow :red : blue is therefore 1:2:4:8.
Once we have the ratio of buttons, we have the probabilities. This is because the probability of picking a blue button is just nb/N, with similar expressions for the other colours. The probabilities are therefore:
- Blue: 8/1+2+4+8 = 8/15
- Red: 4/1+2+4+8 = 4/15
- Yellow: 2/1+2+4+8 = 2/15
- Green: 1/1+2+4+8 = 1/15
Oxford PAT Questions: #2
The given wave is produced by a mechanical oscillator that steadily oscillates the end of the cord at a frequency of 5.0 Hz. The wave has a wavelength of 0.60 metres and an amplitude of 0.030 metres. Draw arrows to indicate the directions in which points Y and Z are moving.
a) Y vertically down, Z vertically down
b) Y vertically down, Z vertically up
c) Z vertically u, Y vertically up
d) Y vertically up, Z vertically down
e) None of the above.
Read the question carefully, then take your time to think before checking the solution.
SOLUTION #2:
To solve this problem, we first need to understand the properties of the wave. Given that the wave has a wavelength of 0.60 metres and an amplitude of 0.030 metres, we can draw the wave profile.
Now, to determine the directions in which points Y and Z are moving, we need to consider the motion of particles in the wave.
For a transverse wave like the one described, particles oscillate perpendicular to the direction of wave propagation. When the wave is at its peak (crest), the particles are moving upwards, and when the wave is at its trough, the particles are moving downwards.
- At point Y, which is at the peak of the wave (crest), the particles are moving upwards.
- At point Z, which is at the trough of the wave, the particles are moving downwards.
Therefore, the correct answer is option d. You also have to draw an upward arrow at point Y and a downward arrow at point Z to represent the directions in which the particles at these points are moving.
Oxford PAT Questions: #3
A ball of mass 0.1 kg bounces on a hard surface. Every time it hits the floor, it loses a quarter of its kinetic energy. If the ball is released from a height of 1.00 m, after how many bounces will the ball bounce no higher than 0.25 m?
Hint:
Remember that gravitational potential energy depends linearly on height above the ground. After each bounce, 75% of the energy remains, so how many times do you need to take 75% of 75% of 75% . . . etc. to have only 25% remaining?
SOLUTION #3:
Firstly, it is worth noticing that the principle of conservation of energy (whereby PE + KE = constant) does not apply in this case since the ball is losing energy after every bounce.
Let the ball start at an initial height h0. Its initial gravitational potential energy is then GPE = mgh0. At the moment before it bounces for the first time, this potential energy has been entirely converted into kinetic energy, such that KE0 = mgh0. However, after its first bounce the ball has lost a quarter of its kinetic energy, such that
KE1 = ¾ mgh0 = mgh1 (1.46)
where h1 is the height the ball reaches after the first bounce and KE1 is its kinetic energy after the first bounce. This can be further simplified into an equation linking initial/final heights:
¾ h0 = h1 (1.47)
and so we see that the mass of the ball makes no difference. Moreover, if the question were changed so that the ball was 10 kg or 2718 kg but everything else remained the same, the answer would still be the same.
Since GPE = mgh, the height, h that the ball reaches depends linearly on the gravitational potential energy. Therefore it will bounce no higher than one quarter of its starting height when it has no more than one quarter of its gravitational potential energy left.
Given that the ball retains ¾ of its energy during each bounce, the height it reaches after n bounces is
hn = (¾ )n × h0 (1.48)
By setting h0 = 1 (since the ball is released from a height of 1.00 m) we can then solve to find when hn ≤ 0.25:
0.25 ≤ (0.75)n (1.49)
ln 0.25 ≤ ln 0.75n (1.50)
ln 0.25 ≤ n ln 0.75 (1.51)
n ≥ 4.81 (1.52)
Oxford PAT Questions: #4
What amount of energy is required to raise a 50 kg student vertically by 0.40m? Assume that all the energy from the hamburger, which has a chemical potential energy of 2.2MJ, can be converted into either mechanical work or potential energy.
- Considering 1MJ equals 1000000 J, calculate how much energy is needed to lift a 50 kg student by 0.40m. Use g = 9.81ms−2 and provide your answer to 2 significant figures.
- How many times must the student perform step-ups of 0.40m to expend the energy equivalent to that of a hamburger? Assume g = 9.81ms−2 and round your answer to 2 significant figures.
Make sure to thoroughly read the question, then allocate time for thoughtful consideration before confirming the solution.
SOLUTION #4:
A. To find the energy needed to lift the student, we can use the formula:
E = mgh
where:
- E is the energy (in joules),
- m is the mass (in kilograms),
- g is the acceleration due to gravity (in metres per second squared),
- h is the height (in metres).
Substituting the given values:
E = (50kg) × (9.81m/s 2 ) × (0.40m)
E = 196.2J
B. To determine how many step-ups the student needs to do to burn off the energy from the hamburger, we then divide the energy of the hamburger by the energy expended in each step-up.
Energy per step-up = mgh
Using the same formula as before, we find the energy per step-up:
Energy per step-up = (50kg) × (9.81m/s 2 ) × (0.40m)
Then, to find the number of step-ups needed:
Number of step-ups = Energy per step-up/Energy of hamburger = 2.2 × 106 J/196.2J
Number of step-ups ≈ 11205
Therefore:
A. The energy needed to lift the student is approximately 196.2J (to 2 significant figures).
B. The student would need to perform approximately 11205 step-ups of 0.40m each to burn off the energy from the hamburger (to 2 significant figures).
Oxford PAT Questions: #5
It takes 20 minutes to fill a bathtub by running the hot water tap; 15 minutes to fill the same bathtub by running the cold water tap; 10 minutes to drain the bathtub by removing the plug. If both taps are running and the plug is removed, how long will it take to fill the bathtub?
SOLUTION #5
Recognising that this is a rate problem, the rate at which the hot water tap fills up the bathtub is 1/20; the cold water tap rate is 1/15; and removing the plug drains the bath at a rate of 1/10. Therefore, when both taps are running and the plug is removed:
rate = 1/time = 1/20 + 1/15 − 1/10 (1.1)
= 3/60 + 4/60 − 6/60 (1.2)
= 1/60 (1.3)
This means that the bath takes 60 minutes to fill.
Oxford PAT Questions: #6
In a dark room against a white wall, you create shadows of your hand using two coloured lamps, one red and one green, as illustrated. The colours of the wall and the shadows on the left and right will respectively appear,
a) Yellow, red, and green
b) Yellow, green, and red
c) Red, green, and yellow
d) Green, red, and yellow
e) Red, yellow, and green
f) None of these
SOLUTION #6:
Answer:
a – yellow, red, and green
Reason:
Firstly, the background appears yellow, a combination of red and green. The left shadow is cast by the green lamp and would appear black if the red lamp were turned off. Therefore, because light from the red lamp falls on it, it is red. Similarly, the right shadow is cast by the red light, and is illuminated with light from the green lamp, so it is green.
Oxford PAT Questions: #7
A pot holding 1.00kg of water is warmed from beneath by the heating component of an electric stove (Figure 1). The environment is at a surrounding temperature of 20.0 ∘C.
Water’s boiling point is at 100 degrees Celsius. When the pot is filled with water at its boiling point, the efficiency of energy transfer from the heating element to the pot is 70.0%. The heated pot then releases energy to the surroundings at a rate described by the formula:
P = kΔT
Where k = 0.760WK−1 and ΔT represents the temperature difference between the pot and the surroundings.
Water’s vaporisation entails a latent heat of 2260 kJ-1 per kilogram. If the water boils at a rate exceeding 0.500 grams per second, the pot overflows, spilling water onto the stove.
- What is the lowest power required to be supplied to the heating element to maintain the water in the pot at its boiling point?
- What is the highest power that can be applied to the heating element, preventing the pot from boiling over?
SOLUTION #7:
A) To find the lowest power required to maintain the water in the pot at its boiling point, we need to calculate the rate at which energy is lost to the surroundings and then determine the power needed to offset this loss.
Given:
- Mass of water (m) = 1.00 kg
- Efficiency of energy transfer (η) = 70.0%
- Latent heat of vaporisation (L) = 2260 kJ/kg
- Surrounding temperature (T_surroundings) = 20.0°C
First, we find the energy lost to the surroundings using the formula:
P = kΔT
Where:
P is the power (in watts)
k is the constant (0.760 W/K)
ΔT is the temperature difference between the pot and the surroundings
ΔT = 100°C − 20°C = 80°C
Now, plug in the values to find P:
P = 0.760 × 80 = 60.8W
This is the minimum power required to maintain the water in the pot at its boiling point.
B) To find the highest power that can be applied to the heating element without the pot boiling over, we need to consider the rate of boiling. So, if the water boils at a rate exceeding 0.500 grams per second, the pot will overflow.
Given: Rate of boiling exceeding = 0.500 g/s
We need to find the power corresponding to this rate of boiling. First, we convert the rate of boiling to kilograms per second:
Rate of boiling = 0.500g/s = 0.0005kg/s
The power required to vaporise the water at this rate can be found using the formula:
P = L × Rate of boiling/η
Substitute the values:
Power (P) = 02260 × 0.0005/0.7
P = 1130/0.70
P ≈ 1614.29W
So, the highest power that can be applied to the heating element without the pot boiling over is approximately 1614.29 watts.
Oxford PAT Questions: #8
Jocko, weighing 60 kg and stationary on ice, catches a 20 kg ball hurled towards him at a velocity of 10 km/h. How fast does Jocko and the ball move over the ice?
SOLUTION #8:
The momentum before the catch is all in the ball, 20 kg x 10 km/h = 200 kg km/h
This is also the momentum after the catch, where the moving mass is 80 kg
…. 60 kg for Jocko and 20 kg for the caught ball.
Thus momentum = 200 kg km/h / 80 kg
= 2.5 km/h
Oxford PAT Questions: #9
Saturn’s second-largest moon is Rhea. Its mass is 2.3 × 1021 kg. It orbits Saturn at a speed of 8.5 × 103 ms−1. Now calculate the kinetic energy of Rhea.
SOLUTION #9:
To calculate the kinetic energy (KE) of Rhea, we use the formula:
KE = ½ × m × v2
Where,
m = 2.3 × 1021 kg
v = 8.5 × 103 ms−1
Plugging in the values:
KE = ½ × (2.3 × 1021) × (8.5 × 103 ms−1)
KE= 1.0375×1030J
So, the kinetic energy of Rhea is 1.0375 × 1030 joules.
Oxford PAT Questions: #10
There are 3 cards hidden under a cloth on a table. It is known that one card is white on both sides, one is black on both sides and the other is black on one side and white on the other.
I select a card at random and its upper face is white – what are the odds that its other side is also white?
Hint:
If you’re familiar with the classic Monty Hall problem, similar thinking should help here. You may think that you have an equal chance of picking any of the three cards – this was true, but looking does more than just eliminate the black card. Furthermore, it may help to label each side from 1 to 6 and assign odds that way.
SOLUTION #10:
Let the white-white card have sides W1 and W2, the black-black card have sides B1 and B2 and the white-black card have sides W3 and B3. With no knowledge other than the fact that I have picked a white side, the side that I have picked could either be W1, W2 or W3:
- If it is W1 then the other side is also white
- Or, if it is W2 then the other side is also white
- If it is W3 then the other side is black
So the probability that the other side is also white is 2/3.
There are a couple of other ways of thinking about the problem:
- It is more likely that I have selected the white-white card since half the time the white-black card will have its black face up
- Consider running the game twice, with all cards flipped the second time around. Removing the cloth, there are three white cards. It is equally likely that you have selected either of these. Moreover, of these, two are the white-white cards. So there is a 2/3 chance that the other side is also white.
A Few Last Words
We hope that you found these 10 Oxford PAT Questions useful and that you had fun solving the physics problems.
We want to make sure you are well and truly prepared for your upcoming PAT test. That’s why we’ve also compiled free notes and PAT past papers with Worked Solutions and Answers! As well as practising PAT questions, you can boost your chances of success with our PAT Tutoring programme.
