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O-Level Dynamics
What the O-Level syllabus expects for Dynamics, and how to practise it.
What the syllabus expects
- Recognise and tell apart contact forces (such as friction, air resistance, tension and normal force) from non-contact forces (such as gravitational, electrostatic and magnetic forces).
- State that a body's mass measures how much matter it contains.
- State that a gravitational field is any region where a mass feels a pull owing to gravitational attraction.
- Define gravitational field strength, g, as the gravitational force per unit mass at a point.
- Recall and use weight = mass × gravitational field strength in unfamiliar situations or problem solving.
- Tell the difference between mass and weight.
- Use Newton's Laws to describe how balanced and unbalanced forces affect a body, how a force can alter a body's motion, and to pick out action-reaction pairs on two interacting bodies.
Scope: Reciting the statements of Newton's Laws is not required. - Recognise the forces on a body and draw free body diagrams showing them, limited to forces acting in at most two dimensions.
- Use a graphical method to solve problems involving a stationary point mass acted on by three forces in two dimensions.
- Recall and use resultant force = mass × acceleration in unfamiliar situations or problem solving.
- Understand that mass is the property giving a body its resistance to changes in motion, known as inertia.
- Explain how friction influences a body's motion.
- Describe how bodies of fixed mass fall through a uniform gravitational field with and without air resistance, mentioning terminal velocity.
How it's examined
Questions on this topic most often ask you to state, calculate, describe, find. About 9% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (2 marks)
Table 14.1 records a skydiver's velocity over a 10 s span; her speed climbs to a maximum of 55 m/s at t = 5 s and afterwards settles to a steady 12 m/s. Using the idea of forces, describe how she arrives at terminal velocity from the moment her parachute becomes fully open at t = 5 s.
Show the worked answer
At t = 5 s the parachute opens, so the air resistance suddenly becomes much larger and exceeds her weight. The resultant force is now upward, so she decelerates. As her speed falls, the air resistance decreases. She keeps slowing until the air resistance has fallen to equal her weight; the resultant force is then zero, so she continues at a constant speed - terminal velocity (12 m/s).
Example 2 (3 marks)
For the 25 g pellet (impact speed 56 m/s, penetration 0.12 m, stopping time 20 ms), calculate its deceleration and hence the resistive force acting on it.
Show the worked answer
Step 1 - state the principle for the deceleration. For motion with uniform acceleration, acceleration = change in velocity / time taken a = (v − u)/t where u is the initial velocity (m/s), v is the final velocity (m/s) and t is the time taken (s). Step 2 - list the quantities in SI units before substituting. u = 56 m/s (the impact speed) v = 0 m/s (the pellet is brought to rest) t = 20 ms = 20/1000 s = 0.020 s Step 3 - substitute into a = (v − u)/t. a = (0 − 56) / 0.020 a = −56 / 0.020 a = −2800 m/s² The minus sign only records that the acceleration points opposite to the motion, so deceleration = 2800 m/s². Step 4 - state Newton's second law to get the force. resultant force = mass × acceleration F = ma with m in kilograms and a in m/s², which gives F in newtons. Step 5 - convert the mass to kilograms. m = 25 g = 25/1000 kg = 0.025 kg Step 6 - substitute, carrying the units through. F = 0.025 kg × 2800 m/s² F = 70 kg m/s² and since 1 N = 1 kg m/s², F = 70 N. This is the resistive force the target exerts on the pellet, acting opposite to the pellet's motion.
Example 3 (4 marks)
Two strings hanging from a rod support a stone. One string carries a tension of 1.3 N and rises at 40 degrees to the horizontal on one side; the other carries 2.0 N and rises at 60 degrees to the horizontal on the opposite side. Using the space given, construct a labelled scale vector diagram (state your scale) and use it to find the magnitude of the resultant force.
Show the worked answer
Step 1 - choose and state a scale. The forces are only a few newtons, so use 1 cm represents 0.2 N That makes the two tensions 1.3 N -> 1.3/0.2 = 6.5 cm 2.0 N -> 2.0/0.2 = 10.0 cm Step 2 - work out the angle between the two forces before you draw. One string rises at 40 degrees above the horizontal on one side and the other at 60 degrees above the horizontal on the opposite side, so measuring round from one to the other through the vertical: angle between them = 180 - 40 - 60 = 80 degrees Step 3 - draw the first vector. Draw a 6.5 cm line at 40 degrees to the horizontal, arrow pointing up along that string, and label it 1.3 N. Step 4 - add the second vector head to tail. From the head of that line, draw a 10.0 cm line at 60 degrees to the horizontal sloping the other way (so it is 80 degrees round from the first), arrow pointing up along the second string, and label it 2.0 N. Step 5 - draw the resultant. Join the tail of the first vector to the head of the second. Put a double arrowhead on it to show it is the resultant. Step 6 - measure that closing line and convert with the scale: length measured ≈ 13 cm resultant = 13 × 0.2 ≈ 2.6 N It comes out vertical, pointing upwards, which makes sense - it is what holds the stone up. Check by calculation: resultant = sqrt(1.3² + 2.0² + 2 × 1.3 × 2.0 × cos 80 degrees) = sqrt(6.59) = 2.6 N Resultant ≈ 2.6 N (2.4 N to 2.8 N accepted), vertically upwards.
More O-Level Pure Physics topics
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