Year 11 Paper 2: Physics Practice
Covers energy, electricity, particle model, forces and cell biology.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [2 marks]
Cell Biology
Prokaryotic cells and eukaryotic cells are the two basic types of cell.
State two differences between a prokaryotic cell and a eukaryotic cell.
Question 2 [2 marks]
Particle Model of Matter
The specific heat capacity of aluminium is 900 J/kg degrees C.
Calculate the energy needed to raise the temperature of 2.0 kg of aluminium by 10 degrees C. Use energy = mass x specific heat capacity x temperature change.
Question 3 [3 marks]
Energy
Energy is conserved in any change involving energy transfer.
State what is meant by the conservation of energy, and describe what eventually happens to energy that is dissipated (wasted).
Question 4 [3 marks]
Forces
A runner travels at an average speed of 4.0 m/s.
Calculate the time it takes the runner to travel 600 m. Use speed = distance / time.
Question 5 [3 marks]
Particle Model of Matter
A gas has a much lower density than the same substance in its solid state.
Explain, in terms of the particle model, why a gas has a much lower density than the same substance as a solid.
Question 6 [4 marks]
Electricity
A resistor at constant temperature and a filament lamp are both circuit components whose resistance can be investigated.
Describe how the resistance of a resistor at constant temperature changes as the current through it increases, and describe how the resistance of a filament lamp changes as the current through it increases.
Question 7 [5 marks]
Cell Biology
An enzyme-controlled reaction is carried out at a range of temperatures from 10 degrees C to 60 degrees C. The enzyme's optimum temperature is 37 degrees C.
Describe and explain how the rate of this enzyme-controlled reaction changes as the temperature increases from 10 degrees C to 60 degrees C.
Question 8 [4 marks]
Forces
A box of mass 20 kg is pulled along the ground by a force of 90 N.
A frictional force of 30 N acts on the box in the opposite direction.
Calculate the acceleration of the box. Use F = m a.
Question 9 [5 marks]
Energy
A weightlifter raises a 120 kg barbell through a height of 2.0 m in 1.5 s.
Calculate the work done raising the barbell, and calculate the power developed by the weightlifter.
Question 10 [4 marks]
Forces
A resultant force of 600 N acts on a car of mass 1500 kg.
Calculate the acceleration of the car. Use F = m a.
Question 11 [5 marks]
Particle Model of Matter
A rectangular block of wood measures 20 cm by 10 cm by 5 cm and has a mass of 750 g.
Calculate the density of the wood, in kg/m^3.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| a eukaryotic cell having a nucleus, while a prokaryotic cell has no nucleus (genetic material free in the cytoplasm) | B1 |
| a eukaryotic cell being much larger than a prokaryotic cell | B1 |
| Final answer: Eukaryotic cells have a nucleus and are larger; prokaryotic cells have no nucleus and are much smaller | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| substituting 2.0 x 900 x 10 | M1 |
| 18000 J | A1 |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| stating energy cannot be created or destroyed | B1 |
| stating energy can only be transferred, stored or dissipated, from one store to another | B1 |
| stating dissipated energy spreads out to the surroundings, becoming increasingly difficult to use for further useful work | B1 |
| Final answer: Energy cannot be created or destroyed, only transferred between stores; dissipated energy spreads out and becomes harder to use usefully | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging speed = distance / time to time = distance / speed | M1 |
| substituting 600 / 4.0 | M1 |
| 150 s | A1 |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| particles in a gas being much further apart than in a solid | B1 |
| the same mass of substance occupying a much larger volume as a gas | B1 |
| density = mass / volume, so a much larger volume for the same mass gives a much lower density | B1 |
| Final answer: Gas particles are much further apart, so the same mass occupies a much larger volume, giving a much lower density | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| the resistance of a resistor at constant temperature staying constant as the current changes | B1 |
| the resistance of a filament lamp increasing as the current through it increases | B1 |
| stating this is because the temperature of the filament increases as more current flows | B1 |
| stating the resistance of a metal increases as its temperature increases, making it harder for the charge carriers to pass through | B1 |
| Final answer: A resistor's resistance stays constant with current; a filament lamp's resistance increases with current, because it heats up and a hotter metal has higher resistance | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| the rate of reaction increasing as temperature increases from 10 degrees C towards 37 degrees C | B1 |
| this being because the enzyme and substrate particles have more kinetic energy and collide more frequently | B1 |
| the rate being at its maximum at the optimum temperature, 37 degrees C | B1 |
| the rate decreasing rapidly above 37 degrees C because the enzyme becomes denatured | B1 |
| stating that denaturing changes the shape of the enzyme's active site, so the substrate no longer fits and the enzyme stops working | B1 |
| Final answer: Rate increases up to the optimum at 37 degrees C as collisions increase, then falls rapidly as the enzyme denatures and its active site changes shape | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| resultant force = 90 - 30 | M1 |
| a resultant force of 60 N | A1 |
| rearranging F = m a to a = F / m, substituting 60 / 20 | M1 |
| 3.0 m/s^2 | A1 |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| using work done = GPE = m g h | M1 |
| substituting 120 x 9.8 x 2.0 | M1 |
| 2352 J | A1 |
| power = work done / time, 2352 / 1.5 | M1 |
| 1568 W | A1 |
| Final answer: 2352 J of work done; 1568 W of power | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to a = F / m | M1 |
| substituting 600 / 1500 | M1 |
| 0.4 | A1 |
| the unit m/s^2 | B1 |
| Final answer: 0.4 m/s^2 | |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| volume = 20 x 10 x 5 | M1 |
| 1000 cm^3 | A1 |
| converting mass to kg (0.75 kg) and volume to m^3 (0.001 m^3) | M1 |
| density = mass / volume, 0.75 / 0.001 | M1 |
| 750 kg/m^3 | A1 |