Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Weight and motion under gravity. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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.