AQA A level maths, Specimen Paper 2: interactive.

Apply Newton's second law: the sum total of all the forces acting on a particle, in any particular direction, is always equal to the force with which the particle is moving: the rate of change of its momentum, which (if the mass is constant) is ma. That produces an equation (4 = 0.05a) which you can solve for a.

Hover the mouse over the pair of rotatable moments, and left click and drag when you are over a grey sausage shape near the join. Not actually at the join, where you'll see a grey circle appear. (Clicking there may change the shape, but if so just press the control key with Z to undo.)

Drag the moments so the join is at the point C in the diagram. Rotate the moments to get a sense of how they depend on both the radius (distance from C) and the force acting perpendicularly at that distance.

See how equilibrium means you can set both equal to each other, and solve for M.

(a) Acceleration is the gradient of the graph as it is the rate of increase in velocity. And since the gradient is constant you can calculate by dividing the increase (V - U) by the time taken (T).

(b) Now rearrange to give an expression for T (in terms of V, U and a) that you can substitute for T in the given formula for S. Then multiply through be 2a and continue making V² the subject.

(a) Observe from the combination of vectors F₁, F₂ and F₃ (far right) that the combination of all the horizontal components is the sum of 2 relatively large ones pointing East, minus one small one opposed to them, resulting in the relatively large horizontal component labelled 25 - 7 + 15 (top); whereas the two upward vertical components opposed to the larger downward one result in the relatively short downward component labelled 12 + 5 - 28. Hence the calculations for (i) the magnitude and (ii) the angle of the resultant, as shown.

(b) Click and drag to compare F₄ with the resultant of the other three, to see how it cancels them out.

(a) Since displacement is the integral function of velocity, the total distance travelled is the value of the total area between the graph and time axis.

(b) Now use the fact that the second of the two areas (from t = 10 to 20) was a distance travelled in the reverse of the original direction, back towards the start.

(c) Using Newton's second law: the resultant of all the forces acting on the particle (the train) is equal to "ma", which is a handy expression for the force with which it is actually moving i.e. its rate of change in momentum. We know the magnitude of this rate will be a maximum when a is at its maximum magnitude which is 2 (the deceleration between t = 6 and t = 10). So F, the resultant (retarding) force is equal to ma which is 800 times 2.

(d) The graph assumes several instantaneous step changes in acceleration.

(a) Even though the velocity vector is changing (both in magnitude and direction) with time, we can represent it with arrows for each component, as shown, top. To get a sense of the dynamic story, hold 'shift' while left click-and-dragging the end of the (red) v arrow, bringing it to a horizontal starting position (representing a horizontal velocity of 40), and then drag to show the increasing vertical component. (We've swapped the signs of the vertical terms to show their downward magnitudes.)

Since velocity is the gradient (derivative) function of displacement, integrate each component of the velocity to find the displacement function s for that component. Horizontally the c value must be 200 so that the position is the origin at time t = 0; vertically a similar reasoning gives c = -250.

(b) Just set the horizontal displacement expression to a value of 100, and balancing steps will solve for t as shown. Then plug that value into the vertical expressions to get the vertical position for that value of t.

(c) As acceleration isn't constant, we need to differentiate the (vertical) velocity function (with respect to t). Then find the value of that derivative function at time t = 0, which is the only time that there is no air resistance involved.

(a) The plan is to use equilibrium (zero velocity vertically but also constant velocity horizontally) to find both the vertical reaction R and the friction F_max, and hence μ which is defined as F_max/R.

See how the 8g weight vector equals the sum of the opposing forces which are the reaction (to be found) and the vertical component (sin40°) of the tension. So you can subtract appropriately to find R, as shown.

Horizontally there are only 2 forces: F_max (to be found) and equal to that (because of constant velocity) the horizontal component (cos40°) of the tension. Finally, divide F_max/R for μ.

(b)(i) Always a good idea anyway to start by doing as here asked: collect all the forces acting in their 'original' directions (before resolving into components) and without considering magnitudes. Weight is still vertical of course. The other 3 vectors rotated 5° as shown.

(b)(ii) See how the equations for vertical and horizontal can be adjusted to become perpendicular and parallel (to the slope), respectively. For perpendicular, the only adjustment needed is finding the appropriate component (cos5°) of the weight.

Parallel, the (2) differences now are (1) a non-zero acceleration, so that Newton's second law (ΣF = ma) requires us to represent ma itself as a vector, as shown; and (2) the involvement of a small component (sin5°) of the weight. Hence, ma equals whatever is left of the T cos40° after subtraction of the two retarding forces.

(a) First find the horizontal component of the initial velocity, which is constant, so using s = ut. Then similarly for the vertical component but including the deceleration due to gravity, so s = ut + ½at².

Then apply v² = u² + 2as to find the velocity at the distance of 3; and finally Pythagoras to calculate the resultant velocity at that instant.

(b) At max height, the vertical velocity is zero and we just have the horizontal velocity of 16.

AQA A level maths 7357 Specimen Paper 3: interactive! (with mouse)

Hover the mouse over a vector for guidance.A Recommended Resource