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Newton's second law, \(F = ma\): the resultant force on a particle equals its mass times its acceleration. You use it in one dimension (forces along and perpendicular to the motion) and in vector form, with forces given in \(\mathbf{i}\), \(\mathbf{j}\) form, usually together with the constant acceleration formulae.
Key facts
\(F = ma\): resultant force = mass × acceleration
1 N gives a mass of 1 kg an acceleration of 1 m s−2
Constant force ⇒ constant acceleration ⇒ suvat applies
The acceleration is in the direction of the resultant force
Notes
Newton's second law
diagram
Newton's second law: resultant force = mass × acceleration, \(F = ma\), with \(F\) in N, \(m\) in kg and \(a\) in m s−2.
The acceleration is in the direction of the resultant force.
Write the equation in the direction of the acceleration: (forces in that direction) − (forces against it) \(= ma\)
Resultant force in the direction of the acceleration = ma
Perpendicular to the motion there is no acceleration, so the forces balance; e.g. on level ground with no other vertical forces, \(R = mg\)
If \(a = 0\) the forces balance (Newton's first law): a car moving at constant speed has driving force = resistance.
Forces and motion together
A constant resultant force gives a constant acceleration, so suvat can be used.
Either forces → \(a\) → suvat (to find speeds, times and distances), or suvat → \(a\) → forces (to find an unknown force).
When a force changes (an engine is switched off, a rope breaks), start a new stage with a new acceleration. The final velocity of one stage is the initial velocity of the next.
Vector form
diagram
With forces as vectors: \(\mathbf{F} = m\mathbf{a}\), where \(\mathbf{F}\) is the vector sum of the forces.
\(\mathbf{a} = \dfrac{1}{m}\mathbf{F}\): divide each component by \(m\). The acceleration is parallel to the resultant force.
The acceleration is parallel to the resultant force
The magnitude of the acceleration is \(|\mathbf{a}|\). With constant forces, use vector suvat: \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\), \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \dfrac{1}{2}\mathbf{a}t^2\)
To find an unknown force: find \(\mathbf{a}\) (e.g. \(\mathbf{a} = \dfrac{\mathbf{v} - \mathbf{u}}{t}\)), work out \(m\mathbf{a}\), then subtract the forces you know.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the acceleration from the forces
3 to 5 marksGrade E
Draw a force diagram and mark the direction of motion.
Resultant force = forward forces − backward forces.
Use \(F = ma\), then suvat if speeds or times are asked for.
Example. A car of mass 1200 kg moves along a straight horizontal road. The engine produces a constant driving force of 3000 N and the total resistance to motion is a constant 600 N. (a) Find the acceleration of the car. (b) The car starts from rest. Find its speed after 5 s.
Use suvat with the given speeds, distances or times to find \(a\)
A negative \(a\) means a deceleration.
Use \(F = ma\) with the resultant force to find the unknown.
Example. A particle of mass 0.5 kg moves in a straight line on a horizontal surface. Its speed decreases from 12 m s−1 to 4 m s−1 while it travels 16 m. The only horizontal force on the particle is a constant resistance of magnitude \(R\) newtons. Find \(R\)
Show the model answerHide the model answer
\(4^2 = 12^2 + 2a(16)\) (M1), \(a = -4\) m s−2
(A1)
\(-R = 0.5 \times (-4)\) (M1), so \(R = 2\)
(A1)
Acceleration from forces in vector form
5 to 7 marksGrade C
Add the force vectors to get the resultant.
Divide by the mass to get \(\mathbf{a}\)
Use \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) for the velocity, and magnitudes for speeds.
Example. A particle \(P\) of mass 2 kg is acted on by two constant forces, \((4\mathbf{i} - 3\mathbf{j})\) N and \((2\mathbf{i} + 11\mathbf{j})\) N, and no others. (a) Find the magnitude of the acceleration of \(P\). (b) \(P\) starts from rest. Find the speed of \(P\) after 3 s.
Show the model answerHide the model answer
(a) Resultant \(= (6\mathbf{i} + 8\mathbf{j})\) N (M1); \(\mathbf{a} = \dfrac{1}{2}(6\mathbf{i} + 8\mathbf{j}) = (3\mathbf{i} + 4\mathbf{j})\) m s−2 (M1 A1); \(|\mathbf{a}| = 5\) m s−2
Find the acceleration vector from the change in velocity: \(\mathbf{a} = \dfrac{\mathbf{v} - \mathbf{u}}{t}\)
Resultant force \(= m\mathbf{a}\)
Unknown force = resultant − known forces; find its magnitude if asked.
Example. A particle of mass 4 kg moves under the action of two constant forces, \(\mathbf{F}_1 = (2\mathbf{i} + 5\mathbf{j})\) N and \(\mathbf{F}_2\). The particle starts from rest and 2 seconds later its velocity is \((6\mathbf{i} - 4\mathbf{j})\) m s−1. Find the magnitude of \(\mathbf{F}_2\)
Using one force instead of the resultant force in \(F = ma\)
Forgetting the resistance or the friction.
Writing \(F = mg\) instead of \(F = ma\)
Adding the magnitudes of forces instead of adding the forces as vectors.
Using a mass in grams, or giving a weight in kg.
More tips
Memory tricks
Write Newton's second law with a direction arrow, e.g. 'N2L (→): 3000 − 600 = 1200a', so the signs are clear.
A particle that starts from rest under a constant force moves in a straight line in the direction of that force.
If \(a\) comes out negative, the particle is decelerating in your positive direction: check that this makes sense.
Exam technique
State the law and the direction each time: method marks are given for a correct N2L equation with the right number of terms.
Give an acceleration as a vector unless the question asks for its magnitude.
Keep full calculator values between parts and round only the final answer.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Apply \(F = ma\) in a straight lineThe resultant force in the direction of motion equals mass × acceleration.
Grade D
Combine \(F = ma\) with suvatFind \(a\) from the forces and then use suvat, or find \(a\) from suvat and then find a force.
Grade C
Use \(F = ma\) with forces as vectorsAdd the force vectors and divide by the mass to get the acceleration vector.
Grade B
Find an unknown force from vector motionFind \(\mathbf{a}\) from the motion, work out the resultant force \(m\mathbf{a}\), then subtract the known forces.
Grade A
Solve problems where the forces changeWhen a force is removed or changes, find the new acceleration and link the stages by the velocity.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
Two forces, \((3\mathbf{i} + 4\mathbf{j})\) N and \((5\mathbf{i} - 10\mathbf{j})\) N, act on a particle \(P\) of mass 2 kg. No other forces act on \(P\). A student writes: The magnitudes of the forces are 5 N and 11.18 N, so the resultant force is 16.18 N and the acceleration is \(16.18 \div 2 = 8.09\) m s−2.
(a) Explain the error in the student’s working.[1]
(b) Find the acceleration of \(P\), giving your answer as a vector.[2]
(c) Find the magnitude of the acceleration of \(P\).[1]
(d) Find the angle between the acceleration of \(P\) and the vector \(\mathbf{i}\).[1]
Show the answer and mark scheme
(a)Answer: Forces are vectors: they must be added as vectors (by components). Magnitudes only add when the forces act in the same direction.
B1 for the forces must be added as vectors (the magnitudes can only be added if the forces are in the same direction)
Worked solution: Forces are vectors, so the resultant is found by adding the components. Adding magnitudes is only correct when the forces act in the same direction, which these do not.
(b)Answer: \((4\mathbf{i} - 3\mathbf{j})\) m s−2
M1 for the resultant \((8\mathbf{i} - 6\mathbf{j})\) N
A1 for \(\mathbf{a} = (4\mathbf{i} - 3\mathbf{j})\) m s−2
Worked solution: Resultant \(= (3 + 5)\mathbf{i} + (4 - 10)\mathbf{j} = (8\mathbf{i} - 6\mathbf{j})\) N. \(\mathbf{a} = \frac{1}{2}(8\mathbf{i} - 6\mathbf{j}) = (4\mathbf{i} - 3\mathbf{j})\) m s−2.
(c)Answer: 5 m s−2
B1 for 5 (m s−2)
Worked solution: \(|\mathbf{a}| = \sqrt{4^2 + 3^2} = 5\) m s−2.
B1 for \(36.9^\circ\) (from \(\tan^{-1}\frac{3}{4}\))
Worked solution: \(\tan\theta = \frac{3}{4}\), so the acceleration makes an angle of \(36.9^\circ\) with \(\mathbf{i}\), below the \(\mathbf{i}\) direction.
Question 2Medium11 marks
A boat of mass 800 kg is pulled from rest across a lake by a horizontal rope. The boat moves in a straight line with constant acceleration and reaches a speed of 3 m s−1 after 12 s. The resistance to the motion of the boat is modelled as a constant force of magnitude 200 N. The rope is modelled as light and inextensible.
(a) Find the acceleration of the boat.[2]
(b) Find the tension in the rope.[3]
(c) When the speed of the boat is 3 m s−1, the rope breaks. Find the distance travelled by the boat after the rope breaks.[4]
(d) In practice, the boat travels further than 18 m after the rope breaks. Identify the modelling assumption that is most likely to be responsible, and explain your answer.[2]
Show the answer and mark scheme
(a)Answer: 0.25 m s−2
M1 for \(3 = 0 + 12a\)
A1 for 0.25
Worked solution: \(v = u + at\): \(3 = 12a\), so \(a = 0.25\) m s−2
(b)Answer: 400 N
M1 for Newton’s second law with the tension and the resistance
A1ft for \(T - 200 = 800 \times 0.25\)
A1 for 400 (N)
Worked solution: \(T - 200 = 800 \times 0.25 = 200\), so \(T = 400\) N
(c)Answer: 18 m
M1 for Newton’s second law with the resistance only
A1 for deceleration \(\frac{200}{800} = 0.25\) m s−2
M1 for \(0 = 3^2 - 2 \times 0.25 \times s\)
A1 for 18 (m)
Worked solution: After the rope breaks: \(200 = 800a\), so the deceleration is \(0.25\) m s−2. \(0 = 9 - 0.5s\), so \(s = 18\) m.
(d)Answer: The resistance is assumed constant. In reality water resistance decreases as the boat slows down, so the deceleration is smaller and the boat travels further.
B1 for identifying the assumption that the resistance is constant
B1 for explaining that the resistance decreases as the speed decreases, so the boat decelerates less and travels further
Worked solution: The resistance of the water depends on the speed of the boat. As the boat slows, the resistance falls below 200 N, the deceleration falls, and the boat travels further than the constant-resistance model predicts.
Question 3Hard10 marks
A car moves along a straight horizontal road. The resistance to the motion of the car is modelled as a constant force of magnitude \(R\) newtons. When the engine produces a driving force of 1670 N, the car accelerates at 1.2 m s−2. When the engine produces a driving force of 900 N, the car accelerates at 0.5 m s−2.
(a) Find the mass of the car and the value of \(R\).[6]
(b) The car is moving at 14 m s−1 when the engine is switched off. Find the distance the car travels before coming to rest.[4]
Show the answer and mark scheme
(a)Answer: Mass 1100 kg; \(R = 350\)
M1 for Newton’s second law for the first case
A1 for \(1670 - R = 1.2m\)
M1 for Newton’s second law for the second case
A1 for \(900 - R = 0.5m\)
dM1 for solving simultaneously (dependent on both M marks)
A1 for mass 1100 kg and \(R = 350\)
Worked solution: \(1670 - R = 1.2m\) and \(900 - R = 0.5m\). Subtracting: \(770 = 0.7m\), so \(m = 1100\) kg and \(R = 1670 - 1320 = 350\).
(b)Answer: 308 m
M1 for Newton’s second law with the resistance only: \(350 = 1100a\)
A1ft for \(a = \frac{7}{22}\) (deceleration)
M1 for \(0 = 14^2 - 2as\) with their \(a\)
A1 for 308 (m)
Worked solution: Deceleration \(= \frac{350}{1100} = \frac{7}{22}\) m s−2. \(0 = 14^2 - 2 \times \frac{7}{22} \times s\), so \(s = 308\) m.