3 exam-style questions plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Weight is the force of gravity on a mass, \(W = mg\), where \(g\) is taken as 9.8 m s−2 unless you are told otherwise. This subtopic covers vertical motion under gravity using suvat (objects thrown up or dropped) and vertical motion with other forces, such as a crate lifted by a cable.
Key facts
\(W = mg\), in newtons
\(g = 9.8\) m s−2 unless told otherwise
Moving freely under gravity: acceleration \(g\) downwards
Greatest height: \(v = 0\)
Up positive: \(a = -9.8\)
Lifting with tension \(T\): \(T - mg = ma\)
Falling with resistance \(R\): \(mg - R = ma\)
\(g\) depends on location: it is not a universal constant
Notes
Weight and \(g\)
Weight is the gravitational force on an object: \(W = mg\) newtons, acting vertically downwards.
\(g\) is the acceleration due to gravity. Use \(g = 9.8\) m s−2 unless the question gives a different value, such as 9.81 or 10.
\(g\) is not a universal constant: it varies slightly with location on the Earth (for example with height and latitude) and is much smaller on the Moon. Near the Earth's surface the model takes it as constant.
Mass stays the same wherever the object is; weight depends on \(g\)
Moving freely under gravity
diagram
'Moving freely under gravity' means the only force is the weight, so the acceleration is \(g\) downwards, whatever the mass.
Choose a positive direction. With up as positive, \(a = -9.8\) and a point below the start has a negative displacement.
With up as positive, a = −9.8 and the ground is at s = −2
At the greatest height \(v = 0\)
Without air resistance the motion is symmetrical: the time up equals the time down to the same level, and the speed at any height is the same going up as coming down.
Vertical motion with other forces
diagram
If other forces act, use \(F = ma\) vertically. For a crate lifted by a cable with tension \(T\): \(T - mg = ma\) (up positive).
Up positive: T − mg = ma
A falling object with air resistance \(R\):\(mg - R = ma\), so its acceleration is less than \(g\)
Accelerating upwards needs \(T > mg\); moving at constant speed needs \(T = mg\); slowing down while moving up needs \(T < mg\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, dM = depends on the earlier method mark.
Vertical projection: greatest height and time to land
5 to 7 marksGrade D
State your positive direction (usually up), so \(a = -9.8\)
Greatest height: \(v = 0\) in \(v^2 = u^2 + 2as\); add the launch height to get the height above the ground.
Time to land: the displacement is minus the launch height; solve the quadratic and take the positive root.
Example. A ball is thrown vertically upwards with speed 14.7 m s−1 from a point 2 m above horizontal ground. The ball is modelled as a particle moving freely under gravity. (a) Find the greatest height of the ball above the ground. (b) Find the time taken for the ball to reach the ground.
Show the model answerHide the model answer
(a) Up positive: \(0 = 14.7^2 - 2(9.8)s\) (M1),\(s = 11.025\) (A1); height above the ground = 13 m (2 s.f.) or 13.0 m (3 s.f.)
(A1)
(b) \(-2 = 14.7t - 4.9t^2\) (M1 A1), so \(4.9t^2 - 14.7t - 2 = 0\) and \(t = \dfrac{14.7 + \sqrt{14.7^2 + 4(4.9)(2)}}{9.8}\) (dM1) = 3.1 s (2 s.f.) or 3.13 s (3 s.f.)
(A1)
Vertical motion with a tension or resistance
3 to 5 marksGrade C
Draw the forces: weight \(mg\) down, tension up, resistance opposing the motion.
Write Newton's second law in the direction of the acceleration.
Use suvat for speeds or times if asked.
Example. A crate of mass 50 kg is raised vertically by a light inextensible cable. The tension in the cable is 560 N. (a) Find the acceleration of the crate. (b) The crate starts from rest. Find its speed after 5 s.
Measure both heights from the same point (e.g. the ground) at the same time \(t\)
Write each height in terms of \(t\), taking care with signs.
Set the heights equal and solve; check that both objects are still in the air.
Example. Ball \(A\) is released from rest at the top of a tower, 40 m above horizontal ground. At the same instant, ball \(B\) is projected vertically upwards from the ground directly below \(A\), with speed 20 m s−1. The balls move freely under gravity. Find the time at which the balls collide and their height above the ground at that moment.
Using \(a = +9.8\) when up has been taken as positive.
Forgetting to add the launch height when asked for the height above the ground.
Taking the negative root of a quadratic in \(t\)
Giving weight in kg, or using \(m\) instead of \(mg\) for the weight in \(F = ma\)
Giving too many significant figures: answers using \(g = 9.8\) should be to 2 or 3 s.f.
More tips
Memory tricks
Thrown up with speed \(u\): time to the top \(\dfrac{u}{g}\), greatest height \(\dfrac{u^2}{2g}\) above the launch point.
With up positive, \(s = ut - 4.9t^2\), since \(\dfrac{1}{2} \times 9.8 = 4.9\)
Look for multiples of 9.8: 14.7 = 1.5g, 19.6 = 2g, 24.5 = 2.5g. They often make answers come out neatly.
In a meeting problem the \(4.9t^2\) terms often cancel, because both objects have the same acceleration.
Exam technique
Write 'taking up as positive' (or down) at the start of your working.
Keep multiples of \(g\) exact, such as \(50g\), until the last line.
Know that \(g\) is not a universal constant: it depends on where you are, and the model treats it as constant near the Earth's surface.
Common assumptions to state: the ball is a particle; air resistance is negligible.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Calculate weight using \(W = mg\)Mass in kg, \(g = 9.8\) m s−2, weight in newtons.
Grade E
Use suvat for vertical motion under gravityTake the acceleration as \(g\) downwards and choose up or down as positive.
Grade D
Find the greatest height and time of flightAt the top \(v = 0\); when it lands the displacement is 0 or a known negative value.
Grade C
Apply \(F = ma\) to vertical motionFor an object lifted by a cable, \(T - mg = ma\)
Grade B
Solve problems with two objects under gravityWrite both heights in terms of \(t\), measured from the same point, and equate them.
Grade A
Handle motion starting above the groundUse a negative displacement for a point below the start, solve the quadratic and reject the negative root.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The acceleration due to gravity at the surface of the Earth is not the same everywhere. It is approximately 9.78 m s−2 at the Equator and approximately 9.83 m s−2 at the North Pole. A climber has mass 70 kg.
(a) Using these values, find the difference between the weight of the climber at the North Pole and the weight of the climber at the Equator.[2]
(b) Explain why the mass of the climber is the same at both places.[1]
(c) Explain why mechanics questions state the value of \(g\) to be used.[1]
Worked solution: Weights: \(70 \times 9.83 = 688.1\) N and \(70 \times 9.78 = 684.6\) N. Difference \(= 3.5\) N.
(b)Answer: Mass is the amount of matter in the climber, which does not depend on location; weight (the gravitational force, \(mg\)) changes because \(g\) changes.
B1 for mass does not depend on location (it is the quantity of matter), whereas weight = \(mg\) depends on \(g\)
Worked solution: Mass measures the quantity of matter and does not change; only the weight \(mg\) changes with \(g\).
(c)Answer: \(g\) is not a universal constant: it depends on location, so a value (such as 9.8 m s−2) must be assumed in the model.
B1 for \(g\) varies with location (it is not a universal constant), so the model needs an agreed approximate value
Worked solution: Because \(g\) varies from place to place (and with height), any model must assume a particular value, such as 9.8 m s−2.
Question 2Medium8 marks
A student drops a ball of mass 0.2 kg from rest from a window. The ball falls vertically through a distance of 12 m to the ground. The student measures the time taken for the ball to reach the ground as 1.7 s. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
(a) Using the model that the ball is a particle falling freely under gravity, find the time that the ball should take to reach the ground.[2]
(b) Suggest one reason why the measured time is longer than the time predicted by the model.[1]
(c) Assuming that the ball falls with constant acceleration, use the measured time to find this acceleration.[2]
(d) Hence estimate the magnitude of the average resistance to the motion of the ball, assuming that the resistance is constant.[3]
Show the answer and mark scheme
(a)Answer: 1.56 s (accept 1.6 s) s
M1 for \(12 = \tfrac{1}{2} \times 9.8 \times t^2\)
A1 for 1.56 or 1.6 (s)
Worked solution: \(12 = 4.9t^2\), so \(t = \sqrt{\frac{12}{4.9}} = 1.56\) s (3 s.f.).
(b)Answer: Air resistance acts on the ball, so its acceleration is less than \(g\) (or: the student’s reaction time adds to the measured time).
B1 for air resistance (reduces the acceleration) / human reaction time in starting or stopping the timer
Worked solution: The model ignores air resistance, which reduces the acceleration of the ball; timing errors due to reaction time are also possible.
(c)Answer: 8.30 m s−2
M1 for \(12 = \tfrac{1}{2}a \times 1.7^2\)
A1 for 8.30 (awrt)
Worked solution: \(12 = \tfrac{1}{2}a(1.7)^2\), so \(a = \frac{24}{2.89} = 8.30\) m s−2 (3 s.f.).
(d)Answer: 0.299 N (accept 0.30 N) N
M1 for Newton’s second law with the weight and the resistance: \(0.2g - R = 0.2a\)
A1ft for a correct equation with their acceleration
A1 for 0.299 or 0.30 (N)
Worked solution: \(0.2 \times 9.8 - R = 0.2 \times 8.304\), so \(R = 0.299\) N (3 s.f.).
Question 3Hard9 marks
A set of bathroom scales measures the force exerted on it and displays this force divided by 9.8, as a reading in kilograms. A person of mass \(m\) kg stands on the scales on the horizontal floor of a lift. When the lift is moving upwards with acceleration \(a\) m s−2, the reading on the scales is 66 kg. When the lift is moving upwards with deceleration \(2a\) m s−2, the reading on the scales is 48 kg. The person is modelled as a particle. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
(a) Find the value of \(m\) and the value of \(a\).[6]
(b) State the reading on the scales when the lift moves at constant speed.[1]
(c) The cable of the lift breaks and the lift falls freely. State the reading on the scales while the lift falls freely, explaining your answer.[2]
Show the answer and mark scheme
(a)Answer: \(m = 60, \ a = 0.98\)
B1 for the forces on the scales: \(66g\) N and \(48g\) N (equal to the reactions on the person, by Newton’s third law)
M1 for Newton’s second law for the person in the first case
A1 for \(66g - mg = ma\)
A1 for \(mg - 48g = 2ma\)
dM1 for eliminating one unknown, e.g. \(2 \times (1) - (2)\) giving \(132g + 48g = 3mg\)
A1 for \(m = 60\) and \(a = 0.98\)
Worked solution: The reading \(r\) kg means the force on the scales is \(9.8r\) N, which equals the reaction \(R\) on the person. \(66g - mg = ma\) (1) and \(mg - 48g = 2ma\) (2). \(2 \times (1) - (2)\): \(132g - 2mg - mg + 48g = 0\), so \(3mg = 180g\) and \(m = 60\). Then \(6g = 60a\), so \(a = \frac{g}{10} = 0.98\).
(b)Answer: 60 kg
B1ft for 60 (kg), their \(m\)
Worked solution: With no acceleration, \(R = mg\), so the reading is 60 kg.
(c)Answer: 0 kg: the person and the scales both accelerate downwards at \(g\), so there is no reaction between them. kg
B1 for 0 (kg)
B1 for the person accelerates at \(g\), so \(mg - R = mg\) gives \(R = 0\)
Worked solution: For the person: \(mg - R = mg\), so \(R = 0\). The scales exert no force on the person, and the reading is 0 kg.