AS Paper 4: Mechanics Focus
Covers proof and algebraic methods, trigonometry, differentiation, integration, kinematics, and forces and Newton's laws.
Questions
Question 1 [3 marks]
Trigonometry
Without using a calculator, find the exact value of sin(45 deg) tan(60 deg) + cos(45 deg), giving your answer in the form (sqrt(a) + sqrt(b))/2.
Question 2 [3 marks]
Integration
Evaluate the definite integral of (3x^2 - 2x + 1) with respect to x between x = -1 and x = 2.
Question 3 [4 marks]
Forces and Newton's Laws
A resultant force acts on a body of mass 12 kg, giving it an acceleration of 2.5 m/s^2.
Find the magnitude of the resultant force, and find the time taken for the body to increase its speed from 4 m/s to 19 m/s under this force.
Question 4 [4 marks]
Differentiation
Without using a calculator, find dy/dx for y = 5x^4 - 3/x^2 + 6, writing 3/x^2 as 3x^-2 before differentiating.
Question 5 [4 marks]
Kinematics
A cyclist decelerates uniformly from 15 m/s to rest in 12 seconds.
Find the deceleration, and find the distance travelled while decelerating.
Question 6 [4 marks]
Differentiation
A curve has equation y = (4x - 1)/(x + 2).
Find dy/dx using the quotient rule, and find the gradient of the curve at the point where x = 1.
Question 7 [3 marks]
Proof and Algebraic Methods
The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.
Find f^-1(x) and state its domain.
Question 8 [4 marks]
Differentiation
A curve has equation y = ln(5x^2 + 1).
Find dy/dx using the chain rule, and find the gradient of the curve at the point where x = 2, giving your answer to 3 significant figures.
Question 9 [4 marks]
Integration
A curve passes through the point (0, 5) and satisfies dy/dx = 6e^(2x) - 4.
Find the equation of the curve.
Question 10 [4 marks]
Trigonometry
Using the small angle approximation cos(x) = 1 - (x^2)/2, valid for small values of x in radians, show that (1 - cos(x))/x^2 is approximately 1/2 for small x.
Question 11 [4 marks]
Integration
Without using a calculator, find the integral of x(x^2 + 1)^4 with respect to x, using the substitution u = x^2 + 1 or otherwise.
Question 12 [5 marks]
Proof and Algebraic Methods
Without using a calculator, express 3/(x + 1) - 2/(x - 2) as a single fraction in its simplest form.
Question 13 [5 marks]
Kinematics
A stone is thrown horizontally with speed 15 m/s from the top of a cliff 20 m high. Using g = 9.8 m/s^2,
find the time taken for the stone to reach the ground, and find the horizontal distance it travels before landing.
Question 14 [5 marks]
Proof and Algebraic Methods
Prove that n^2 - n is even for every integer n.
Question 15 [6 marks]
Integration
Use the trapezium rule with 4 strips to estimate the value of the integral of sqrt(1 + x^3) with respect to x between x = 0 and x = 2, giving your answer to 3 significant figures.
Use a table of values at x = 0, 0.5, 1, 1.5, 2.
Question 16 [6 marks]
Kinematics
A ball A is projected vertically upwards from ground level with speed 24.5 m/s. At the same instant, a ball B is dropped from rest from a point 40 m directly above the launch point of A. Using g = 9.8 m/s^2,
find the time at which the two balls collide, and find the height above the ground at which they collide.
Question 17 [6 marks]
Trigonometry
Solve 2 cos(2x) + 3 cos(x) = 0 for 0 <= x <= 360 deg, giving all solutions to 1 decimal place.
Question 18 [6 marks]
Forces and Newton's Laws
A particle of mass 2 kg is held in equilibrium on a rough plane inclined at 30 degrees to the horizontal by a horizontal force of magnitude P newtons, acting in the vertical plane containing the line of greatest slope. The coefficient of friction between the particle and the plane is 0.2, and the particle is on the point of slipping down the plane.
Find the value of P. Use g = 9.8 m/s^2.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| stating sin(45 deg) = sqrt(2)/2, tan(60 deg) = sqrt(3) and cos(45 deg) = sqrt(2)/2 | M1 |
| computing sin(45 deg) tan(60 deg) = sqrt(6)/2 | M1 |
| the total (sqrt(6) + sqrt(2))/2 | A1 |
| Final answer: (sqrt(6) + sqrt(2))/2 | |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative x^3 - x^2 + x | M1 |
| substituting the limits x = 2 and x = -1 | M1 |
| the value 9 | A1 |
| Final answer: 9 | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using F = ma | M1 |
| F = 30 N | A1 |
| using v = u + at to form 19 = 4 + 2.5t | M1 |
| t = 6 s | A1 |
| Final answer: F = 30 N; t = 6 s | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| rewriting 3/x^2 as 3x^-2 | M1 |
| differentiating each term | M1 |
| the 20x^3 term | A1 |
| dy/dx = 20x^3 + 6/x^3 | A1 |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using a = (v - u)/t | M1 |
| a = -1.25 m/s^2 (a deceleration of 1.25 m/s^2) | A1 |
| using s = (u + v)/2 x t | M1 |
| s = 90 m | A1 |
| Final answer: deceleration = 1.25 m/s^2; distance = 90 m | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| applying the quotient rule with u' = 4 and v' = 1 | M1 |
| dy/dx = 9/(x + 2)^2 after simplification | A1 |
| substituting x = 1 | M1 |
| gradient = 1 | A1 |
| Final answer: dy/dx = 9/(x + 2)^2; gradient at x = 1 is 1 | |
| Question 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1 | M1 |
| collecting x terms to give x(y - 2) = -1 - 3y | M1 |
| f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2 | A1 |
| Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the chain rule d/dx[ln(u)] = (1/u)(du/dx) | M1 |
| dy/dx = 10x/(5x^2 + 1) | A1 |
| substituting x = 2 | M1 |
| the gradient = 20/21 (or 0.952 to 3 sf) | A1 |
| Final answer: dy/dx = 10x/(5x^2 + 1); gradient at x = 2 is 20/21 (0.952 to 3 sf) | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| integrating dy/dx = 6e^(2x) - 4 to give y = 3e^(2x) - 4x + c | M1 |
| substituting x = 0 and y = 5 into the equation | M1 |
| the equation 3 + c = 5 | A1 |
| the equation of the curve y = 3e^(2x) - 4x + 2 | A1 |
| Final answer: y = 3e^(2x) - 4x + 2 | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting the approximation cos(x) = 1 - (x^2)/2 | M1 |
| 1 - cos(x) = (x^2)/2 | A1 |
| dividing by x^2 | M1 |
| (1 - cos(x))/x^2 = 1/2, as required | A1 |
| Final answer: (1 - cos(x))/x^2 is approximately 1/2 for small x, since 1 - cos(x) = (x^2)/2 using the small angle approximation | |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the reverse chain rule, using u = x^2 + 1 so du/dx = 2x | M1 |
| rewriting the integral as (1/2) x the integral of u^4 with respect to u | M1 |
| integrating to (1/2)(u^5/5) = u^5/10 | A1 |
| substituting back to (x^2 + 1)^5/10 + c | A1 |
| Final answer: (x^2 + 1)^5/10 + c | |
| Question 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing both fractions over the common denominator (x + 1)(x - 2) | M1 |
| expanding the first numerator to 3(x - 2) = 3x - 6 | M1 |
| expanding the second numerator to 2(x + 1) = 2x + 2 | M1 |
| combining to give the numerator x - 8 | A1 |
| the final answer (x - 8)/((x + 1)(x - 2)) | A1 |
| Final answer: (x - 8)/((x + 1)(x - 2)) | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| using s = 0.5 g t^2 vertically with s = 20 | M1 |
| t = 2.02 s (3 sf) | A1 |
| using the horizontal velocity of 15 m/s, which is unaffected by gravity | M1 |
| horizontal distance = 15 x 2.02 | M1 |
| horizontal distance = 30.3 m (3 sf) | A1 |
| Final answer: t = 2.02 s (3 sf); horizontal distance = 30.3 m (3 sf) | |
| Question 14[5 marks] | |
|---|---|
| Answer or working | Marks |
| factorising n^2 - n as n(n - 1) | M1 |
| identifying that n and n - 1 are two consecutive integers | M1 |
| stating that exactly one of two consecutive integers must be even | A1 |
| using this to deduce that the product n(n - 1) is a multiple of 2 | M1 |
| concluding that n^2 - n is even for every integer n, completing the proof | A1 |
| Final answer: Proof: n^2 - n = n(n - 1), the product of two consecutive integers, one of which is always even, so n^2 - n is always even. | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| evaluating y at x = 0, 0.5, 1, 1.5, 2 (values 1, 1.061, 1.414, 2.092, 3, to 3 dp or better) | M1 |
| identifying the strip width h = 0.5 | M1 |
| using the trapezium rule formula (h/2)[y0 + y4 + 2(y1 + y2 + y3)] | M1 |
| substituting to (0.25)[4 + 2(4.567)] | A1 |
| evaluating the bracket to 13.133 | A1 |
| the final estimate 3.28 (3 sf) | A1 |
| Final answer: Integral (approx) = 3.28 (3 sf) | |
| Question 16[6 marks] | |
|---|---|
| Answer or working | Marks |
| the height of A above the ground: h_A = 24.5t - 4.9t^2 | M1 |
| the height of B above the ground: h_B = 40 - 4.9t^2 | M1 |
| setting h_A = h_B and simplifying to 24.5t = 40 | M1 |
| t = 1.63 s (3 sf) | A1 |
| substituting t into h_B = 40 - 4.9t^2 | M1 |
| height = 26.9 m (3 sf) | A1 |
| Final answer: t = 1.63 s (3 sf); height above the ground = 26.9 m (3 sf) | |
| Question 17[6 marks] | |
|---|---|
| Answer or working | Marks |
| using cos(2x) = 2 cos^2(x) - 1 | M1 |
| substituting and rearranging to 4 cos^2(x) + 3 cos(x) - 2 = 0 | M1 |
| solving using the quadratic formula to cos(x) = (-3 + sqrt(41))/8, rejecting the root outside [-1, 1] | A1 |
| cos(x) = 0.425 (3 sf) | A1 |
| using inverse cosine and the symmetry x = 360 - x | M1 |
| x = 64.8 deg or x = 295.2 deg (1 dp) | A1 |
| Question 18[6 marks] | |
|---|---|
| Answer or working | Marks |
| resolving perpendicular to the plane: R = mg cos(30) + P sin(30) | M1 |
| resolving along the plane with friction F = mu R acting up the slope (point of slipping down): P cos(30) + F = mg sin(30) | M1 |
| substituting F = 0.2R to form P cos(30) + 0.2(mg cos(30) + P sin(30)) = mg sin(30) | A1 |
| collecting terms in P to give P(cos(30) + 0.2 sin(30)) = mg(sin(30) - 0.2 cos(30)) | M1 |
| P = mg(sin(30) - 0.2 cos(30)) / (cos(30) + 0.2 sin(30)) | A1 |
| P = 6.63 N (3 sf) | A1 |