P17: Newton's Laws and Momentum
Covering Newton's three laws of motion, F=ma, inertia, momentum, conservation of momentum, the relationship between force and momentum, and impact forces.
Newton's First Law
Newton's first law states that if the resultant force on an object is zero, the object will remain at rest or continue moving at a constant velocity. If the resultant force is non-zero, the object will accelerate in the direction of the resultant force.
This law describes the concept of inertia:
- An object at rest stays at rest unless a resultant force acts on it
- An object moving at constant velocity keeps moving at that velocity unless a resultant force acts on it
Inertia is the tendency of an object to resist a change in its state of motion. The greater the mass of an object, the greater its inertia - a larger force is needed to change its velocity.
Worked Example
A spacecraft in deep space is travelling at 10 000 m/s with its engines off. There is no friction in space and no resultant force. According to Newton's first law, it will continue at 10 000 m/s indefinitely until a force acts on it.
Newton's Second Law
Force = mass × acceleration
F = ma
F = resultant force (N), m = mass (kg), a = acceleration (m/s²)
Newton's second law states that the acceleration of an object is directly proportional to the resultant force and inversely proportional to the mass of the object.
Worked Example
A car of mass 1200 kg has a resultant force of 3600 N. Calculate its acceleration.
a = F / m = 3600 / 1200 = 3 m/s²
Worked Example
A force of 50 N accelerates a trolley of mass 2 kg. Calculate the acceleration.
a = F / m = 50 / 2 = 25 m/s²
The unit of force, the newton, is defined using Newton's second law:
1 newton is the force needed to accelerate a mass of 1 kg by 1 m/s². Therefore, 1 N = 1 kg × 1 m/s².
In exam questions, always identify the resultant force first by considering all forces acting on the object, then apply F = ma. Common mistakes include forgetting to subtract friction or drag from the driving force.
Newton's Third Law
Newton's third law states that for every action, there is an equal and opposite reaction. Whenever two objects interact, they exert equal and opposite forces on each other.
Important points about Newton's third law pairs:
- The two forces are always the same type (both gravitational, both contact, etc.)
- The two forces act on different objects
- The two forces are equal in magnitude
- The two forces act in opposite directions
Worked Example
A book rests on a table. The book exerts a downward contact force on the table, and the table exerts an equal and opposite upward contact force on the book. These are a Newton's third law pair because they are the same type of force (contact) and act on different objects.
Do not confuse Newton's third law pairs with balanced forces. Weight and the normal contact force on a stationary object are equal and opposite, but they are NOT a Newton's third law pair because they act on the same object and are different types of force.
To identify a Newton's third law pair: both forces must be the same type, act on different objects, and be equal and opposite. If two equal and opposite forces act on the same object, they are simply balanced forces, not a third law pair.
Inertia and Mass (Higher)
The inertial mass of an object is a measure of how difficult it is to change its velocity. Inertial mass can be found using Newton's second law: m = F / a. A larger inertial mass requires a larger force to produce the same acceleration.
Worked Example
Compare the force needed to accelerate a 1000 kg car and a 50 kg go-kart by 2 m/s².
Car: F = 1000 × 2 = 2000 N
Go-kart: F = 50 × 2 = 100 N
The car requires 20 times more force due to its greater inertial mass.
Momentum
Momentum = mass × velocity
p = mv
p = momentum (kg m/s), m = mass (kg), v = velocity (m/s)
Momentum is a vector quantity - it has both magnitude and direction. An object at rest has zero momentum. All moving objects have momentum.
Worked Example
Calculate the momentum of a 1500 kg car travelling at 20 m/s.
p = mv = 1500 × 20 = 30 000 kg m/s
Worked Example
A 0.5 kg ball moves at 8 m/s east. Calculate its momentum.
p = 0.5 × 8 = 4 kg m/s east
Conservation of Momentum (Higher)
In a closed system, the total momentum before an event equals the total momentum after the event. This is the principle of conservation of momentum.
Total momentum before = Total momentum after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Worked Example - Collision
A trolley of mass 3 kg moving at 4 m/s collides with a stationary trolley of mass 2 kg. They stick together. Calculate the combined velocity.
Momentum before = (3 × 4) + (2 × 0) = 12 kg m/s
Momentum after = (3 + 2) × v = 5v
5v = 12, so v = 2.4 m/s
Worked Example - Explosion
A gun of mass 5 kg fires a bullet of mass 0.01 kg at 400 m/s. Calculate the recoil velocity of the gun.
Total momentum before = 0 (both at rest)
Momentum after = 5 × v + 0.01 × 400 = 5v + 4
0 = 5v + 4, so 5v = -4, v = -0.8 m/s (gun recoils at 0.8 m/s in the opposite direction)
Momentum is conserved in all collisions and explosions, provided the system is closed (no external forces act). Kinetic energy may not be conserved - in an inelastic collision, some kinetic energy is transferred to other stores (thermal, sound).
Force and Momentum (Higher)
Force = change in momentum / time
F = (mv - mu) / t
F = force (N), m = mass (kg), v = final velocity (m/s), u = initial velocity (m/s), t = time (s)
This equation shows that the force is equal to the rate of change of momentum. A larger force is needed to produce the same change in momentum in a shorter time.
Worked Example
A ball of mass 0.4 kg hits a wall at 10 m/s and bounces back at 6 m/s. The contact time is 0.2 s. Calculate the force on the ball.
Take the initial direction as positive.
F = (mv - mu) / t = (0.4 × (-6) - 0.4 × 10) / 0.2
F = (-2.4 - 4) / 0.2 = -6.4 / 0.2 = -32 N
The force is 32 N in the opposite direction to the initial motion (away from the wall).
When objects bounce back, remember that the final velocity has the opposite sign to the initial velocity. The change in momentum (mv - mu) will be large because both the magnitude and direction change.
Impact Forces and Safety
In a collision, the change in momentum is fixed. However, by increasing the time of the impact, the force can be reduced. This is because F = Δp / t - increasing t decreases F.
Safety features that increase impact time to reduce force:
- Seat belts - stretch slightly, increasing the time for the body to decelerate
- Airbags - inflate to provide a soft cushion, increasing the stopping time
- Crumple zones - at the front and rear of cars, deform to absorb energy and increase stopping time
- Cycle helmets - compress on impact, increasing stopping time for the head
- Playground surfaces - soft materials cushion falls, increasing stopping time
Worked Example
A 70 kg person in a car travelling at 20 m/s is stopped by an airbag in 0.6 s. Without the airbag, they would hit the dashboard and stop in 0.02 s. Compare the forces.
With airbag: F = (70 × 0 - 70 × 20) / 0.6 = -1400 / 0.6 = -2333 N
Without airbag: F = -1400 / 0.02 = -70 000 N
The airbag reduces the force by a factor of 30.
Questions on impact forces often ask you to explain how a safety feature reduces injuries. Use the chain of reasoning: feature increases impact time → rate of change of momentum decreases → force on person decreases → injuries are less severe.
Required Practical: F = ma
The relationship F = ma can be investigated using a trolley on a runway, with masses on a hanging hook providing the force.
Method:
- Set up a trolley on a runway with a pulley at the end. Attach a hanging mass via a string over the pulley.
- The total mass of the system is the trolley mass plus the hanging mass. Keep the total mass constant by transferring masses from the trolley to the hook.
- Use light gates or a ticker timer to measure the acceleration of the trolley for each value of the force.
- Vary the force by changing the mass on the hook. Record force and acceleration.
- Plot a graph of acceleration against force. The graph should be a straight line through the origin, confirming a ∝ F.
To keep the total mass of the system constant, when you move a mass from the trolley to the hook, the total mass stays the same but the accelerating force increases. This ensures you are only changing the independent variable (force).
Practice Questions
1. State Newton's three laws of motion.
First law: An object remains at rest or continues at constant velocity unless acted on by a resultant force. Second law: F = ma, the acceleration is proportional to force and inversely proportional to mass. Third law: For every action there is an equal and opposite reaction; when two objects interact they exert equal and opposite forces on each other.
2. A resultant force of 150 N acts on a mass of 25 kg. Calculate the acceleration.
a = F/m = 150/25 = 6 m/s².
3. A trolley of mass 4 kg moving at 6 m/s collides with a stationary trolley of mass 3 kg. They stick together. Calculate the velocity after the collision.
Momentum before = 4 × 6 + 3 × 0 = 24 kg m/s. After: (4+3)v = 24, so v = 3.43 m/s.
4. A 60 kg person in a crash decelerates from 15 m/s to rest. The seat belt increases the stopping time to 0.5 s. Calculate the average force exerted by the belt.
F = (60×0 - 60×15)/0.5 = -900/0.5 = -1800 N (1800 N in the opposing direction to motion).
5. Explain, using momentum ideas, why a crumple zone reduces injuries in a crash.
The crumple zone increases the time of the collision. Since the change in momentum is the same, a longer time means a smaller rate of change of momentum, so the force on the occupants is reduced. A smaller force causes less severe injuries.
Required Practical: Investigating F = ma
Aim
To investigate the relationship between force, mass and acceleration, verifying Newton's second law (F = ma).
Equipment
Trolley, runway, pulley, string, mass hanger, slotted masses, light gates and data logger (or ticker timer and ticker tape), metre ruler.
Method
- Set up a trolley on a raised runway with a pulley at one end. Attach a string from the trolley over the pulley to a hanging mass holder.
- Place the light gates at a measured distance apart along the runway. Connect them to a data logger.
- Add slotted masses to the trolley to set the total system mass. Record the mass of the system (trolley + all masses + hanging mass).
- Release the trolley and record the acceleration measured by the light gates.
- To vary force while keeping total mass constant: transfer a mass from the trolley to the hanger (this increases the accelerating force without changing the total system mass). Repeat for at least 6 different forces.
- To vary mass while keeping force constant: add masses only to the trolley (not the hanger). Repeat for at least 6 different masses.
Expected Results
- Graph of acceleration against force: straight line through the origin, confirming a ∝ F (constant mass)
- Graph of acceleration against 1/mass: straight line through the origin, confirming a ∝ 1/m (constant force)
- Gradient of a vs F graph = 1/m (the reciprocal of the total system mass)
Common Errors
Friction on the runway: compensate by tilting the runway slightly so the trolley runs at constant speed without the hanging mass. Not keeping total mass constant when varying force: always transfer masses between trolley and hanger rather than simply adding to the hanger. Parallax errors when reading ticker tape: use light gates instead for greater accuracy.
Maths Skills
F = ma Calculations
Rearrange F = ma to find any unknown: a = F/m, m = F/a. Always identify the resultant force first by considering all forces, then apply the equation. A common exam technique is to combine F = ma with other equations (e.g. weight, friction, kinetic energy).
Momentum: p = mv
Momentum is a vector. Use p = mv for individual objects. For conservation: m1u1 + m2u2 = m1v1 + m2v2. Remember to include signs for direction (e.g. one direction positive, opposite negative).
Force and Momentum: F = (mv − mu)/t
This equation shows that force equals the rate of change of momentum. A larger change in momentum over a shorter time produces a larger force. This is the basis of all car safety features — increasing the impact time reduces the force.
Worked Example
A 1500 kg car travelling at 20 m/s crashes into a barrier and stops in 0.4 s. Calculate the force exerted.
F = (mv − mu)/t = (1500 × 0 − 1500 × 20)/0.4 = −30 000/0.4 = −75 000 N (75 kN opposing motion)
If a crumple zone increases the stopping time to 0.8 s: F = −30 000/0.8 = 37 500 N — the force is halved.
Common Misconceptions
Heavy Objects Falling Faster
Heavier objects always fall faster than lighter ones. In a vacuum (no air resistance), all objects fall at the same rate regardless of mass because a = F/m = mg/m = g. In air, heavier objects often fall faster because air resistance has less effect relative to their weight, but this is due to air resistance, not gravity itself.
Momentum vs Force
Momentum and force are the same thing. Momentum (p = mv) is a property of a moving object — it depends on mass and velocity. Force (F = ma or F = Δp/t) is an interaction that changes momentum. Momentum is what an object has; force is what acts on an object to change what it has. The relationship is that force equals the rate of change of momentum.
6-Mark Extended Question
Use Newton's laws to explain why a seatbelt reduces the force on a passenger during a car crash. [6 marks]
When a car crashes, it decelerates rapidly (1). According to Newton's first law, the passenger continues moving forward at the original speed because there is initially no force to slow them down (1). Without a seatbelt, the passenger would hit the dashboard and stop in a very short time (1). The change in momentum is the same regardless of stopping method: Δp = m × Δv (1). From Newton's second law, F = Δp/t, the force depends on the time over which the momentum changes (1). A seatbelt stretches slightly, increasing the stopping time compared to hitting the dashboard. Since the same change in momentum occurs over a longer time, the force on the passenger is reduced, resulting in less severe injuries (1).
AO3: Analyse and Evaluate
In an experiment, a trolley of mass 0.8 kg moving at 3 m/s collides with a stationary trolley of mass 0.5 kg. After the collision, they stick together and the combined velocity is measured as 2.2 m/s.
(a) Calculate the total momentum before and after the collision. (b) Is momentum conserved? (c) Calculate the kinetic energy before and after the collision. Is kinetic energy conserved? (d) A second trial gives a combined velocity of 1.9 m/s. Calculate the force on the 0.5 kg trolley if the collision lasted 0.15 s.
Evaluation
(a) Before: p = 0.8 × 3 + 0.5 × 0 = 2.4 kg m/s. After: p = (0.8 + 0.5) × 2.2 = 1.3 × 2.2 = 2.86 kg m/s.
(b) The values differ (2.4 vs 2.86), so momentum appears not to be conserved. This suggests experimental error, such as friction on the runway or an inaccurate velocity measurement, since momentum must be conserved in a closed system.
(c) KE before = ½ × 0.8 × 3² = 3.6 J. KE after = ½ × 1.3 × 2.2² = 3.15 J. KE is not conserved (3.6 J → 3.15 J); 0.45 J is transferred to thermal and sound energy, confirming an inelastic collision.
(d) Change in momentum of 0.5 kg trolley = 0.5 × 1.9 − 0 = 0.95 kg m/s. F = Δp/t = 0.95/0.15 = 6.3 N.
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