Short answer questions

Some short answer questions will be multiple choice questions. These will appear in both exam papers, and at both tiers. Multiple choice questions are asked as questions, often starting with 'What is …' or 'Which of these …'.

You have four options to choose from in a multiple choice question. You must only choose one of these options, by writing your answer (A, B, C, or D) in a box.

It may help to reject any answers that you feel are obviously wrong, so that you can focus on choosing the right answer.

Other short answer questions will start with command words such as 'describe' or 'explain'. Some command words are easy to understand such as:

  • 'calculate' or 'determine' for maths questions
  • 'complete' to fill in a gap in a table or graph
  • 'define' to give the meaning of an important word
  • 'suggest' where you use your knowledge in an unfamiliar situation

The command words 'describe' and 'explain' can be confusing. If you are asked to describe a graph, you will be expected to write about its overall shape, whether it is linear or curved, the slope of gradients etc. If you are asked to explain why a pattern or trend is seen in a graph, you will be expected to use your science knowledge, not just say what you see (which is a description), eg 'The graph shows a steep linear increase for the first three hours because…'.

'Explain how' and 'why' questions often have the word 'because' in their answer. 'Describe' questions don't.

The number of marks per question part is given in this form '[2 marks]'. It is essential that you give two different answers if a question is worth two marks. Sometimes you can gain a second mark by giving the units in a calculation or stating specific data points, eg 'The speed of the object decreased by 8m/s.'

Questions courtesy of Eduqas.

Sample question 1 - Foundation

Question

The velocity-time graph shows part of the journey of a train.

Graph plotting time in seconds against velocity in metres per second. Line moves upwards then across and then back down again.

The gradient of the line gives the acceleration of the train. The area under the line gives the distance travelled by the train.

a) Calculate the initial acceleration of the train. [2 marks]

b) Calculate the total distance travelled by the train. [2 marks]

a) Acceleration = gradient of line [1] = 25 ÷ 50 = 0.5 m/s2 [1]

b) Area under graph, split into 3 sections. Two triangles and a rectangle.

½ × 50 × 25 = 625 m

25 × 150 = 3,750 m

½ × 100 × 25 = 1,250 m

[1]

Total distance = 1,250 + 3,750 + 625 = 5,625 m [1]

Sample question 2 - Higher

Question

Two stationary linked space vehicles, A and B are about to separate.

Vehicle A has a mass of 50,000 kg.

Vehicles A and B are at rest before the separation so the total momentum then is zero.

After the separation, vehicle A moves with a velocity of -2 m/s.

a) Recall an equation to calculate the momentum of A after the separation. [2 marks]

b) No momentum is lost when they separate. Write down the momentum of B after they separate. [1 mark]

c) Vehicle B has a mass of 80,000 kg. Use the same equation to find the velocity of vehicle B after the separation. [2 marks]

a) Momentum of A after separation = mass × velocity [1]

= 50,000 × -2 = -100,000 kg m/s [1]

b) Momentum before = momentum after

0 = momentum A + momentum B

momentum B = - momentum A = - (- 100,000) = + 100,000 kgm/s [1]

c) velocity B = momentum B ÷ mass B [1]

= (+ 100,000) ÷ 80,000 = + 1.25 m/s [1]

Sample question 3 - Higher

Question

a) State Newton's first law of motion. [2 marks]

b) i) A car of mass 900 kg experiences an accelerating force of 2,250 N. Recall an equation to calculate its acceleration. [2 marks]

ii) If the car was initially travelling at 5 m/s, recall an equation to calculate its final velocity after 8 s. [2 marks]

a) An object continues in its state of rest/inertia/motion/constant speed in a straight line [1] unless acted upon by an external/unbalanced force [1].

b) i) F  =  ma [1]

a  =  \frac{F}{m} =  \frac{2,250}{900} =  2.5~m/s^2 [1]

ii) a  = (v  -  u)  \div  t [1]

2.5  =  (v  -  5)  \div 8

2.5  \times  8  =  v  -  5

20  +  5  =  v  =  25~m/s [1]