What you'll learn
- State and apply Newton’s three laws of motion.
- Calculate linear momentum using p=mvp = mvp=mv, including direction.
- Use F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp and recognise F=maF = maF=ma as a special case.
- Find impulse from FΔtF\Delta tFΔt and from the area under a force–time graph.
Before Newton’s laws: forces are vectors
A force is a push or pull on an object. Force is measured in newtons, N.
A force is a vector quantity, meaning it has both magnitude and direction. When several forces act on an object, their combined effect depends on direction as well as size.
Resultant force
The resultant force is the single force that has the same effect as all the forces acting on an object added together as vectors. It is also called the net force.
A free-body diagram shows one object by itself, with all the external forces acting on it. This is usually the best starting point for Newton’s laws questions.
Choose a positive direction
For one-dimensional motion, choose one direction as positive. Forces, velocities, momenta and accelerations in the opposite direction are then negative.
Newton’s three laws of motion
Newton’s laws are a model for predicting motion. They connect what you can observe — forces and motion — with quantitative equations that can be tested by experiment.
Newton’s first law
Newton’s first law
An object remains at rest or continues moving with constant velocity unless acted on by a resultant external force.
“Constant velocity” means constant speed and constant direction. So an object moving in a straight line at steady speed has zero resultant force.
This does not mean there are no forces. It means the forces are balanced.
Newton’s second law
Newton’s second law gives the link between resultant force and change in motion. The most general form you need here is:
F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔpwhere FFF is the resultant force, Δp\Delta pΔp is the change in momentum, and Δt\Delta tΔt is the time interval.
For an object of constant mass, this becomes:
F=maF = maF=mawhere mmm is mass in kilograms, kg, and aaa is acceleration in metres per second squared, m s^-2.
Newton’s third law
Newton’s third law
When two bodies interact, they exert forces on each other that are equal in magnitude, opposite in direction, and of the same type.
The two forces act on different objects. For example, if your hand pushes a wall, the wall pushes your hand back with an equal and opposite force.
Cancelling third-law pairs
Third-law force pairs do not cancel each other when analysing one object, because they act on different objects. Only forces acting on the same object can be added to find the resultant force.
Finding acceleration from a resultant force
A 2.0 kg trolley is pulled forwards by 6.0 N. Friction acts backwards with a force of 1.5 N. Find the acceleration.
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Choose forwards as positive and calculate the resultant force:
F=6.0 N−1.5 N=4.5 NF = 6.0\ \text{N} - 1.5\ \text{N} = 4.5\ \text{N}F=6.0 N−1.5 N=4.5 N -
Use Newton’s second law for constant mass:
F=maF = maF=maso
a=Fma = \frac{F}{m}a=mF -
Substitute the values, carrying units through:
a=4.5 N2.0 kg=2.25 m s−2a = \frac{4.5\ \text{N}}{2.0\ \text{kg}} = 2.25\ \text{m s}^{-2}a=2.0 kg4.5 N=2.25 m s−2To two significant figures, the acceleration is 2.3 m s^-2 forwards.
Linear momentum
Linear momentum
The linear momentum ppp of an object is the product of its mass and velocity:
p=mvp = mvp=mvMomentum is measured in kilogram metre per second, kg m s^-1.
Momentum is a vector because velocity is a vector. Its direction is the same as the direction of motion.
If motion is along a straight line, you can use positive and negative signs to show direction.
Momentum has direction
Two objects can have the same speed but opposite momenta if they are moving in opposite directions.
Calculating vector momentum
A 0.145 kg tennis ball travels to the right at 38 m s^-1. Take right as positive. Find its momentum.
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Identify the mass and velocity:
m=0.145 kg,v=+38 m s−1m = 0.145\ \text{kg}, \qquad v = +38\ \text{m s}^{-1}m=0.145 kg,v=+38 m s−1 -
Use the momentum equation:
p=mvp = mvp=mv -
Substitute:
p=0.145 kg×38 m s−1=5.51 kg m s−1p = 0.145\ \text{kg} \times 38\ \text{m s}^{-1} = 5.51\ \text{kg m s}^{-1}p=0.145 kg×38 m s−1=5.51 kg m s−1The momentum is 5.5 kg m s^-1 to the right.
Net force as rate of change of momentum
Newton’s second law is most fundamentally about momentum:
F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔpThis says the resultant force is the rate of change of momentum. A large force produces a rapid change in momentum. The force acts in the direction of the momentum change.
For constant mass:
p=mvΔp=mΔvF=mΔvΔtF=ma\begin{aligned} p &= mv \\ \Delta p &= m\Delta v \\ F &= \frac{m\Delta v}{\Delta t} \\ F &= ma \end{aligned}pΔpFF=mv=mΔv=ΔtmΔv=maSo F=maF = maF=ma is a special case of F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp when the mass is constant.
Use the correct form
For this section, you usually deal with constant mass objects, so F=maF = maF=ma is often valid. But the deeper statement is F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp.
Finding average force from momentum change
A 0.060 kg ball travelling left at 20 m s^-1 is hit so that it leaves travelling right at 15 m s^-1. The contact time is 5.0 ms. Take right as positive. Find the average force on the ball.
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Assign signs to the velocities:
u=−20 m s−1,v=+15 m s−1u = -20\ \text{m s}^{-1}, \qquad v = +15\ \text{m s}^{-1}u=−20 m s−1,v=+15 m s−1 -
Calculate the change in momentum:
Δp=mv−mu=m(v−u)\Delta p = mv - mu = m(v-u)Δp=mv−mu=m(v−u) Δp=0.060 kg×(15−(−20)) m s−1=2.1 kg m s−1\Delta p = 0.060\ \text{kg} \times \left(15 - (-20)\right)\ \text{m s}^{-1} = 2.1\ \text{kg m s}^{-1}Δp=0.060 kg×(15−(−20)) m s−1=2.1 kg m s−1 -
Convert the time into seconds and use F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp:
Δt=5.0 ms=5.0×10−3 s\Delta t = 5.0\ \text{ms} = 5.0 \times 10^{-3}\ \text{s}Δt=5.0 ms=5.0×10−3 s F=2.1 kg m s−15.0×10−3 s=420 NF = \frac{2.1\ \text{kg m s}^{-1}}{5.0 \times 10^{-3}\ \text{s}} = 420\ \text{N}F=5.0×10−3 s2.1 kg m s−1=420 NThe average force is 420 N to the right.
Impulse
Impulse
The impulse of a force is the product of the force and the time interval for which it acts:
impulse=FΔt\text{impulse} = F\Delta timpulse=FΔtImpulse is measured in newton seconds, N s. Since impulse equals change in momentum, N s is equivalent to kg m s^-1.
Impulse is useful for collisions, impacts and explosions, where a force may act for a very short time.
Impulse changes momentum
Impulse is equal to the change in momentum:
FΔt=ΔpF\Delta t = \Delta pFΔt=ΔpA smaller average force can produce the same momentum change if it acts for a longer time.
That is why crumple zones, airbags and padded mats reduce injury risk: they increase the time taken for momentum to change, reducing the average force.
Using impulse to find a speed change
A constant force of 12 N acts on a 0.80 kg object for 0.50 s in the direction of motion. Find the change in velocity.
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Calculate the impulse:
impulse=FΔt=12 N×0.50 s=6.0 N s\text{impulse} = F\Delta t = 12\ \text{N} \times 0.50\ \text{s} = 6.0\ \text{N s}impulse=FΔt=12 N×0.50 s=6.0 N s -
Use impulse equals change in momentum:
Δp=6.0 kg m s−1\Delta p = 6.0\ \text{kg m s}^{-1}Δp=6.0 kg m s−1 -
Since the mass is constant, Δp=mΔv\Delta p = m\Delta vΔp=mΔv:
Δv=Δpm=6.0 kg m s−10.80 kg=7.5 m s−1\Delta v = \frac{\Delta p}{m} = \frac{6.0\ \text{kg m s}^{-1}}{0.80\ \text{kg}} = 7.5\ \text{m s}^{-1}Δv=mΔp=0.80 kg6.0 kg m s−1=7.5 m s−1The object’s velocity increases by 7.5 m s^-1.
Force–time graphs and impulse
For a force–time graph, the area under the graph gives the impulse.
For a constant force, the graph is a rectangle:
area=FΔt\text{area} = F\Delta tarea=FΔtFor a varying force, the graph may be curved. You can estimate the area by counting squares or by splitting it into trapezia. In practical data analysis, a spreadsheet can calculate each strip area and sum them, which is especially useful when many force readings are recorded.
Graph-area units
On a force–time graph, area has units N s, because the axes are force in N and time in s. That immediately tells you the area is an impulse.
Estimating impulse from a force–time graph
A force sensor records these forces at intervals of 0.020 s during a collision: 0 N, 120 N, 180 N, 100 N, 0 N. Estimate the impulse using trapezia.
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The time spacing is constant, so use trapezia between neighbouring readings:
area of one trapezium=12(F1+F2)Δt\text{area of one trapezium} = \frac{1}{2}(F_1 + F_2)\Delta tarea of one trapezium=21(F1+F2)Δt -
Add the four trapezium areas:
impulse≈0.020 s2[(0+120)+(120+180)+(180+100)+(100+0)] N=0.010 s×800 N=8.0 N s\begin{aligned} \text{impulse} &\approx \frac{0.020\ \text{s}}{2} \left[(0+120)+(120+180)+(180+100)+(100+0)\right]\ \text{N} \\ &= 0.010\ \text{s} \times 800\ \text{N} \\ &= 8.0\ \text{N s} \end{aligned}impulse≈20.020 s[(0+120)+(120+180)+(180+100)+(100+0)] N=0.010 s×800 N=8.0 N s -
Interpret the result as a change in momentum:
Δp=8.0 kg m s−1\Delta p = 8.0\ \text{kg m s}^{-1}Δp=8.0 kg m s−1So the object’s momentum changes by about 8.0 kg m s^-1 in the direction of the net force.
Forgetting direction on graphs
If a force–time graph goes below the time axis, that area is negative impulse. Areas above and below the axis can cancel because impulse is a vector quantity.
In the exam
- Start with a free-body diagram and find the resultant force before choosing an equation.
- For momentum and impulse questions, define a positive direction and keep signs for velocities and forces.
- On force–time graphs, use area for impulse: rectangles for constant force, trapezia or square-counting for curved graphs.
Check yourself
- What does Newton’s first law say about an object moving at constant velocity?
- Why is momentum a vector quantity?
- How would you estimate impulse from a curved force–time graph?
