Edexcel A level Maths (9MA0) · Mechanics › Forces and Newton's laws
Practise Newton's second law and vectors. 2 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 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 2850 N, the car accelerates at 1.5 m s−2. When the engine produces a driving force of 1800 N, the car accelerates at 0.8 m s−2.
(a) Find the mass of the car and the value of \(R\).[6]
(b) The car is moving at 10 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 1500 kg; \(R = 600\)
M1 for Newton's second law for the first case
A1 for \(2850 - R = 1.5m\)
M1 for Newton's second law for the second case
A1 for \(1800 - R = 0.8m\)
dM1 for solving simultaneously (dependent on both M marks)
A1 for mass 1500 kg and \(R = 600\)
Worked solution: \(2850 - R = 1.5m\) and \(1800 - R = 0.8m\). Subtracting: \(1050 = 0.7m\), so \(m = 1500\) kg and \(R = 2850 - 2250 = 600\).
(b)Answer: 125 m
M1 for Newton's second law with the resistance only: \(600 = 1500a\)
A1ft for \(a = 0.4\) (deceleration)
M1 for \(0 = 10^2 - 2as\) with their \(a\)
A1 for 125 (m)
Worked solution: Deceleration \(= \frac{600}{1500} = 0.4\) m s−2. \(0 = 10^2 - 2 \times 0.4 \times s\), so \(s = 125\) m.