Physics notes, chapter 2
Force and Motion I
Textbook Physics Form 4 KSSM (DLP), pages 24–75
What this chapter covers
- Distinguish distance, displacement, speed, velocity and acceleration, and use ticker tape and photogates to measure them (2.1)
- Use the four equations of uniformly accelerated linear motion to solve problems (2.1)
- Interpret and sketch displacement-time, velocity-time and acceleration-time graphs, and use gradients and areas (2.2)
- Explain free fall and gravitational acceleration, and solve problems on objects thrown upwards or dropped (2.3)
- Explain inertia and its relationship with mass using Newton's First Law (2.4)
- Calculate momentum and apply the Principle of Conservation of Momentum to collisions and explosions (2.5)
- Use F = ma and Newton's Second Law to solve problems (2.6)
- Explain impulse and impulsive force, with Newton's Third Law, and their uses and safety applications (2.7)
- Distinguish mass and weight and use W = mg and gravitational field strength (2.8)
2.1Linear motion
Linear motion is motion in a straight line. It is described using distance, displacement, speed, velocity and acceleration.
| Distance | Displacement | |
|---|---|---|
| Meaning | Length of the route covered | Shortest distance from the initial to the final position, in a specific direction |
| Type of quantity | Scalar (size only) | Vector (size and direction) |
| Depends on the route? | Yes | No, only on the start and end points |
| Quantity | Formula | S.I. unit |
|---|---|---|
| Speed (scalar) | speed = distance travelled ÷ time taken | m s−1 |
| Velocity (vector) | v = s ÷ t (rate of change of displacement) | m s−1 |
| Acceleration (vector) | a = (v − u) ÷ t (rate of change of velocity) | m s−2 |
Worked example
Dina walks 60 m east, then 20 m back west, in 40 s.Distance = 60 + 20 = 80 m. Displacement = 60 − 20 = 40 m east.
Speed = 80 ÷ 40 = 2 m s−1. Velocity = 40 ÷ 40 = 1 m s−1 east.
A negative acceleration (deceleration) means the velocity is decreasing, or that the acceleration points opposite to the chosen positive direction.
Ticker timer and ticker tape
A ticker timer on a 50 Hz supply makes 50 dots every second, so 1 tick = 1/50 = 0.02 s. A tape pulled by a moving object records both displacement and time.
| Time | Number of ticks |
|---|---|
| 0.02 s | 1 tick |
| 0.10 s | 5 ticks |
| 0.20 s | 10 ticks |
- Dots equally spaced: uniform velocity. Spacing getting larger: acceleration. Spacing getting smaller: deceleration.
- Velocity = length of a strip ÷ (number of ticks × 0.02 s).
- Acceleration = (final velocity − initial velocity) ÷ time between the two strips.
- A photogate and electronic timer is more accurate than a ticker timer: there is no tape friction, and the timer reads to 0.001 s.
Worked example
A 5-tick strip is 3.0 cm long. Time = 5 × 0.02 = 0.10 s, so velocity = 3.0 ÷ 0.10 = 30 cm s−1.Equations of uniformly accelerated motion
v = u + at (1)
s = ½(u + v)t (2)
s = ut + ½at2 (3)
v2 = u2 + 2as (4)
s = ½(u + v)t (2)
s = ut + ½at2 (3)
v2 = u2 + 2as (4)
u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time. Write down the given values, choose the equation that contains the unknown and the quantities you know, then substitute.
Worked example
A cyclist moving at 8 m s−1 brakes and stops after 2 m. Find the acceleration.u = 8, v = 0, s = 2. Use v2 = u2 + 2as: 0 = 64 + 4a, so a = −16 m s−2 (opposite to the motion).
Using an equation when the acceleration is not constant. These four equations only work for uniform acceleration. Also check the units: change km h−1 to m s−1 first (divide by 3.6).
A car accelerates from rest at 3 m s−2 for 6 s. Find its velocity and displacement.
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v = 0 + 3 × 6 = 18 m s−1. s = 0 + ½ × 3 × 62 = 54 m.
2.2Linear motion graphs
| Graph | Gradient gives | Area under graph gives |
|---|---|---|
| Displacement-time | Velocity | (not used) |
| Velocity-time | Acceleration | Displacement |
| Acceleration-time | (not used) | Change in velocity |
- Displacement-time: horizontal line = at rest; straight sloping line = uniform velocity; curve getting steeper = acceleration; downward slope = moving in the negative direction.
- Velocity-time: horizontal line = uniform velocity; sloping straight line = uniform acceleration; below the axis = moving in the negative direction.
For this journey the distance is 100 + 100 + 50 = 250 m but the displacement is −50 m. Average speed = 250 ÷ 20 = 12.5 m s−1; average velocity = −50 ÷ 20 = −2.5 m s−1.
Split the area under a v–t graph into triangles and rectangles, or use a trapezium: area = ½ (sum of parallel sides) × width. Area below the time axis is negative displacement.
Reading the height of a displacement-time graph as velocity. Velocity is the gradient, not the height.
What does a horizontal line on a displacement-time graph show? What does it show on a velocity-time graph?
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Displacement-time: the object is at rest (velocity zero). Velocity-time: uniform velocity (acceleration zero).
2.3Free fall motion
Free fall is motion where the only force acting on an object is the gravitational force (no air resistance or friction). All objects then fall with the same gravitational acceleration, g.
- In air, a feather falls slower than a steel ball because of air resistance; in a vacuum they reach the floor together.
- g = 9.81 m s−2 on the Earth's surface in calculations. It is 9.78 at the equator and 9.83 at the poles because the Earth is not a perfect sphere (the poles are nearer the centre).
- In an experiment with two photogates, g = 2h ÷ (t22 − t12).
Use the linear motion equations with a = g. Take upward as positive, so g = −9.81 m s−2 for an object going up or down.
Worked example
A ball is thrown upwards at 10 m s−1. At the highest point v = 0.Time up: 0 = 10 − 9.81t, so t = 1.02 s. Maximum height: 0 = 102 + 2(−9.81)s, so s = 5.10 m.
Worked example
A stone is dropped from a 10 m cliff. −10 = ½(−9.81)t2, so t = 1.43 s. v2 = 2(−9.81)(−10), so |v| = 14.0 m s−1 downwards.A light hollow plastic ball and a solid steel ball of the same size are dropped from a cliff. Do they land together?
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No. Air resistance is a larger fraction of the weight of the light ball, so it accelerates less and lands later. In a vacuum they would land together.
2.4Inertia
Inertia is the tendency of an object to remain at rest or, if moving, to continue moving in a straight line at uniform velocity.
Newton's First Law: an object remains at rest or moves at uniform velocity unless acted on by an external force.
Newton's First Law: an object remains at rest or moves at uniform velocity unless acted on by an external force.
- Inertia and mass: the larger the mass, the larger the inertia. An inertial balance shows this: a larger mass oscillates with a longer period. It is used to measure mass in zero gravity (astronauts on the ISS).
- Effects: passengers lurch forward when a bus brakes; raindrops fly off a stopped umbrella; sauce leaves a bottle when it is jerked and stopped; tankers have separate tanks so the fuel does not surge; seat belts hold passengers back.
Writing "inertia is a force". Inertia is a property of matter (tendency). It is not a force and has no unit.
Why do passengers in a car move forward when the car stops suddenly?
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Their inertia keeps them in their state of motion (moving forward) while the car stops.
2.5Momentum
Momentum, p = mv (mass × velocity). It is a vector in the direction of the velocity. S.I. unit: kg m s−1 (= N s).
The Principle of Conservation of Momentum: in a closed system with no external force, the total momentum before a collision equals the total momentum after it.
m1u1 + m2u2 = m1v1 + m2v2
- Collision: bodies that stick together move with a common velocity: m1u1 + m2u2 = (m1 + m2)v.
- Explosion: an object at rest breaks into parts: 0 = m1v1 + m2v2, so the parts move in opposite directions (rockets, guns recoiling).
Worked example
A 2 kg trolley at 3 m s−1 hits a 1 kg trolley at rest and they join. Total momentum before = 2 × 3 = 6 kg m s−1, so 6 = (2 + 1)v and v = 2 m s−1.Worked example (explosion)
A gun of mass 2.0 kg fires a 10 g bullet at 100 m s−1. 0 = 0.010(100) + 2.0v, so the gun recoils at v = −0.5 m s−1.A 1 500 kg car moves at 12 m s−1. What is its momentum?
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p = 1 500 × 12 = 18 000 kg m s−1.
2.6Force
- At constant mass, acceleration is directly proportional to the force: a ∝ F (straight line through the origin on an a–F graph).
- At constant force, acceleration is inversely proportional to mass: a ∝ 1/m (straight line through the origin on an a–1/m graph).
Newton's Second Law: the rate of change of momentum is directly proportional to the force and acts in the direction of the force.
F = ma 1 N is the force that gives a 1 kg mass an acceleration of 1 m s−2.
F = ma 1 N is the force that gives a 1 kg mass an acceleration of 1 m s−2.
Worked example
A 1 200 kg car at 30 m s−1 stops in 5 s. a = (0 − 30) ÷ 5 = −6 m s−2. F = 1 200 × (−6) = −7 200 N; the force acts opposite to the motion.Using F = ma for the net force only. If friction or other forces act, use the resultant force: Fnet = ma.
2.7Impulse and impulsive force
Impulse is the change of momentum: J = mv − mu = Ft. Unit: N s.
Impulsive force is the rate of change of momentum in a collision over a short time: F = (mv − mu) ÷ t.
Impulsive force is the rate of change of momentum in a collision over a short time: F = (mv − mu) ÷ t.
For the same change of momentum, F ∝ 1/t. A short impact time gives a large force; a long impact time gives a small force.
| Aim | How | Example |
|---|---|---|
| Large impulse | Follow-through (longer contact) | Kicking a football, golf swing |
| Small force | Longer impact time | Crumple zone, airbag, bending the knees on landing, cushion |
| Large force | Shorter impact time | Hammer, pestle on a mortar |
Newton's Third Law: for every action there is a reaction of equal magnitude in the opposite direction. The action and reaction act on different bodies (a rower pushes water back, water pushes the boat forward).
Worked example
A 0.058 kg plasticine ball hits a wall at 10 m s−1 and sticks. J = 0.058(0) − 0.058(10) = −0.58 N s.2.8Weight
Weight is the gravitational force on an object: W = mg, in newtons, towards the centre of the Earth. Gravitational field strength, g = W ÷ m, unit N kg−1; on Earth g = 9.81 N kg−1.
| Mass | Weight | |
|---|---|---|
| Meaning | Amount of matter | Gravitational force on the object |
| Type | Scalar | Vector |
| Unit | kg | N |
| Changes with place? | No | Yes (depends on g) |
Worked example
The Moon's field strength is one sixth of the Earth's. A 60 kg astronaut weighs 588.6 N on Earth and 588.6 ÷ 6 = 98.1 N on the Moon. The mass is still 60 kg.A 10 kg object weighs 150 N on a planet. What is g there?
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g = W ÷ m = 150 ÷ 10 = 15 N kg−1 (stronger than Earth's 9.81).