Topic - force as rate of change of momentum; (i) force on a wall due to water from a hose, (ii) force on a table as a rope is dropped onto it.

The question starts with a simple calculation that illustrates how a force can be calculated as the rate of change of momentum. It should be understood by most students; however, the answer can then be used in the next part. By splitting up the second, harder calculation into stages it should resemble the style of questions prevalent in A2 past papers and provide a memorable example of the topic.(i) The answer to this part can be deduced by considering the momentum of the amount of water that strikes the wall in a time t, which is p=dAv²t for density d, cross-sectional area A and speed v. It hence follows that the force is F=dAv². The student should explain their working; if they struggle, I can talk them through the necessary steps.(ii) The rope in question, with density d and cross sectional area A, is released from rest with its lower end just touching the table. By realising that the rope is in free fall, they should understand that its speed at a time t is v=g*t and that the length on the table is l=0.5gt². We also have v²=2gl. These are examples of 'suvat' equations of motion, another standard topic in A-level physics. Then, the student should realise that the falling rope leads to a force dAv²=dAg²t² on the table as in (i) due to its momentum changing. However, it should also be noted that the weight of the rope on the table leads to an additional force dAlg, meaning the total force is F=1.5dAg²t² at a time t. The final follow-up question is to use the expression for F to deduce the maximum force on the table which can be shown to be 3mg, where m is the mass of the rope. This surprising result can easily be demonstrated using a chain and some scales. The question must be broken-up into sections to make it accessible to most students.