UPDATE: Papers and mark schemes are now available from the maths papers and physics papers pages.
June 2013 papers will not be accessible from this website until they are made publicly available by the exam boards. However, I can send you a link for the June 2013 Edexcel Physics and Maths A-level papers and mark schemes if you enter your email below.
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Find the current through each resistor in the circuit below:
The solution of this problem requires the use of Kirchhoff’s Laws. Read more ›
Evaluate |AB X CD| where A is (6, -3, 0), B is (3, -7, 1), C is (3, 7, -1) and D is (4,5,-3). Hence find the shortest distance between AB and CD
The two key points two remember here are:
Thus the vector representing the shortest distance between AB and CD will be in the same direction as (AB X CD), which can be written as a constant times (AB X CD). We can then equate this vector to a general vector between two points on AB and CD, which can be obtained from the vector equations of the two lines. Solving the three equations obtained simultaneously, we can find the constant and the shortest distance.
Here is my solution: Read more ›
A circular ring has radius 1cm and is suspended horizontally by a thread passing through the 5cm mark on a ruler pivoted at its centre. The ring is balanced by a 5g mass suspended at the 60cm mark of the ruler.A beaker containing a liquid is then placed under the ring such that it just touches the liquid’s surface horizontally. If the 5g mass is moved to the 70cm mark, the ring ‘just parts’ the surface. Determine the surface tension of the liquid.
We first find the mass of the ring using principle of moments (sum of moments is zero for the balanced ruler) Then we use the same principle for the case when there is a moment from the surface tension of the liquid to find what the surface tension is.
Here is my solution:
Prove that
\( \sum_{r=1}^n rr! = (n+1)!-1 \)
for \( n>1\)
In order to prove a mathematical identity, one needs to show that the identity is valid for all the values in the desired range. Trying all the values in that range is practically impossible.
The induction method starts by proving the identity for the smallest value in the range, n=1 in this case.
It then assumes that the identity is true for another value, usually taken as n=k. Based on that assumption, it proves that the identity is true for the next value, n=k+1.
Given the identity is true for n=1, this shows it is also true for n=2, which shows it is true for n=3 and so on…
The steps to follow for a proof by induction are:
1) Prove for n=1
2) Assume true for n=k
3) Prove for n=k+1
Below you can find my solution for the question above
This is a conceptual question that was asked me about the energy of a trolley attached to a spring, moving on a frictionless horizontal surface.
Let’s say there’s a wall, with a spring attached to it, and a trolley attached to the spring. And the ground the trolley moves on is frictionless, and no air resistance etc… when I pull the spring to one direction, it’ll gain strain energy, yet they’ll be restoring forces acting on it trying to return it to its original position. So on release, the elastic potential energy is lost because the extension is decreasing, but since energy is conserved it’s converted to Kinetic energy.. so E.P.E is decreasing –> K.E is increasing, hence the trolley gains momentum, until eventually the spring goes back to its original shape where the restoring force stops acting and no net force acts and the trolley is at its max speed/velocity, then it continues to move forward because… it still has momentum? and by newtons first law, it’ll keep moving in a straight line.. and since it still has kinetic energy, it moves forward.. but as the spring compresses again it loses it in the form of strain energy.. and that happens until all k.e. –> strain energy and it comes momentarily to rest and then restoring force is dominant and it moves backwards and oscillates to and fro forever…
But if that’s true, why doesn’t the trolley move backwards right away, the moment that the restoring force acts on it? what keeps it moving forward? I’m quite confused in that.. Is it the trolley’s inertia?
Visit our new page for the Maths A-level from 2017 solutionbanks
I have uploaded solution banks for all the Edexcel Heinemann textbooks. They are in pdf format so can be opened without Internet Explorer.
