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  1. S

    ---> Physics Paper 4 Past Paper Discussion <---

    For 3 b) 1, I guess you could just average out the maximum and minimum potential across the resistor, because the minimum is 4.8 V on the graph and the maximum is 6 V on the graph. Taking the average, it gives us 5.4 V, which should be the right way of getting the value (at least I hope so!)...
  2. S

    ---> Physics Paper 4 Past Paper Discussion <---

    Damn, damn, damn, so, so sorry - which question did I reply on? I'm just really confused now, I just checked my earlier post, and it concerns a question you asked about 08, and I replied about 07 - sorry!! Are you good on that one or still out there for that question? Really, really, sorry!
  3. S

    ---> Physics Paper 4 Past Paper Discussion <---

    Sorry, about that, didn't see it.... Okay, so for part 1. of the question, the frequency should remain constant; damping in a system does not change the frequency of the system; the process of damping basically reduces the speed/ velocity of the oscillation of the system; when this happens...
  4. S

    Some help here please for A2 physics!!!

    The basic idea of temperature is that it is a measure of the kinetic energy component of a substances internal energy, as this equation shows: (Image taken from http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html) Going down that page, you can find a more visible form of this...
  5. S

    ---> Physics Paper 4 Past Paper Discussion <---

    The total energy of an oscillating system during the said oscillations remains constant; i) At the maximum displacement the kinetic energy goes down to zero, but potential energy is stored in whatever object(s) provide the force that acts towards the center of the motion (e.g. energy is stored...
  6. S

    ---> Physics Paper 4 Past Paper Discussion <---

    For this question, the requirement is to calculate the angular speed ω, and to do this, there are some things we need to realize first before diving into the calculations: i) The elastic cord stretches whenever there is tension in it. Now, when the cord is rotating, it's pretty difficult to...
  7. S

    Some help here please for A2 physics!!!

    The idea of temperature is that thermal energy will be transferred from a body/object/system with a higher temperature to a body/object/system of a lower temperature, i.e. thermal energy flows from a body with high temperature to a body with a lower temperature. If body A and body B were at...
  8. S

    M1 ON 13 MAY ,WHO HAS THE BEST PREPATION,Post your doubts here?

    For this part, it is necessary to find out how long A is in the air until the time when it reaches a height of 1.3 meters above the ground for the second time. When B has fallen 0.65 meters in 0.5 seconds, A is already at a height of 1.3 meters above the ground level. Since the string is...
  9. S

    Mechanics M1: Post your doubt here

    The total work done on the ball while it's going up and coming done can be related to the change in it's kinetic energy according to the work - kinetic energy theorem: The change in K.E. while going upwards: 0.5 * 0.6 * (0^2 - 5.2^2) = - 8.112 J Therefore, the total work done on the ball...
  10. S

    Mechanics M1: Post your doubt here

    When the direction changes, the value of the velocity is negative; you then need to get the value of t when the velocity slows to zero and changes direction. You can get this time by setting v=0 in the given equation to get, 0.75t^2 − 0.0625t^3 = 0 t^2(0.75 - 0.0625t) = 0 Dividing both...
  11. S

    Mechanics M1: Post your doubt here

    Sorry I didn't answer your initial question, I got ahead of myself:( I'm pretty sure they mean that the work done by the force resisting motion on the journey from A to B was 360 kJ; the work done by the force resisting motion on the journey from B to C should be an entirely different value...
  12. S

    Mechanics M1: Post your doubt here

    The work done by the resistive force is the magnitude of the resistive force multiplied by the distance the force moves through; for the frictional/resistive force, this distance is the hypotenuse of each triangle. The hypotenuse of the left - hand side triangle is given by: sin(5) = 45/hyp...
  13. S

    Mechanics M1: Post your doubt here

    Sorry I couldn't reply, had to have lunch ;)
  14. S

    M1 ON 13 MAY ,WHO HAS THE BEST PREPATION,Post your doubts here?

    According to what the marking scheme says, the graph is more like an upward slope, then an inverted mountain: If t = 0, the initial velocity = 5, so the graph must start at t = 0 and v = 5. After that, from t till T1, the question states that the velocity is increasing. This can be shown on...
  15. S

    Mechanics M1: Post your doubt here

    Sure, here it is: Since the system is in equilibrium and just about to move, the friction applied by the rod on the ring is limiting friction. Since the coefficient of friction is 0.4, the frictional force = 0.4 * Normal Force = 0.4 * (Tcos65 + 40) = 0.169T + 16 - that's where they get the...
  16. S

    Mechanics M1: Post your doubt here

    You need to take a reference point in all these cases; by saying that you've taken "t" as time for Particle P, you're defining that t = 0 when P has just been thrown. Imagine the situation; P is thrown at time t = 0, and 0.4 seconds later, Q is thrown upwards. After time t passes, the variable...
  17. S

    M1 ON 13 MAY ,WHO HAS THE BEST PREPATION,Post your doubts here?

    This is just a link to another thread, all credit goes to qffdhruba: https://www.xtremepapers.com/community/threads/mechanics-m1-post-your-doubt-here.20348/page-20 It's the first post on the page. Good Luck for all your exams!
  18. S

    Mechanics M1: Post your doubt here

    Okay, so for part (ii), you are again told that the system is in equilibrium, but that the ring is about to slide up the rod. Friction tries to prevent any relative motion between two surfaces, so if the ring is going to move upwards relative to the rod, the friction exerted BY the rod ON the...
  19. S

    Mechanics M1: Post your doubt here

    Man, is this a tough question! The examiner's report has made it clear that this question bugged students the most, but anyways, here's a stab at it: Since we know the system is in equilibrium, the vector sum of the forces at any point in the system (any point on the rope, on the rod, on the...
  20. S

    Mechanics M1: Post your doubt here

    For 5 (i), you can consider each one of the two blocks separately, A separately and B separately. Since the system is given to be in equilibrium, the acceleration of every part (A and B) of the system will be zero. Thus, Newton's Second Law becomes Net Force = m x a = m x 0 = 0 We can...
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