Physics Past Year Chapter 17

From StpmWiki

Jump to: navigation, search

PAPER 1

2009

23. If work W\, is required to move a charge q\, from point X\, to point Y\,, the potential difference between X\, and Y\, is Media:2009-1-23.jpg

  • A Wq^{2}\,
  • B Wq\,
  • C \frac{W}{q}
  • D \frac{W}{q^{2}}
  • Suggested Answer C
  • Suggested Solution
    • V=\frac{W}{q}\qquad\qquad \mbox{(definition)}
  • Suggested Explanation
    • Physical quantities involved : work, potential difference
    • What is definition of potential difference? Work done .... to bring a unit charge the charge from one point to the other against the electric field
    • V=\frac{W}{q}\qquad\qquad \mbox{(definition)}
  • Alternative Methods / Further investigation /Related questions
    • Using formula W=qV\,
      • I won't encourage the method of using W=qV\, in this case, as the formula comes from the definition, not the other way round, unless you are fully aware of where the formula comes from. What you MUST NOT do is try to memorize this formula blindly.
    • Checking Units/ Dimension
      • Checking units (see which answer has the same unit as the question) is sometimes promoted as a "trick" way to get an answer, but it is actually quite a worthless trick. Time is better spent understanding things the proper way. And after all, it doesn't work here unless you realize that V=JC-1 (Which will require understanding that potential is work done per unit charge anyway)
    • What about the work required, point X to point Y and between X and Y?
      • That is how potential difference is defined, as it is always between two points, it is work done to bring a unit charge from one point to the other point against the electric field
      • The information stated actually could be important if positive/negative sign is important in this question, but it is not.
      • It could be important in comparison to potential, which refers to one point, is the work done to bring a unit charge from infinity to that point
    • Revision/Related Info
      • Potential, potential difference Work done ... per unit charge ...
      • Potential energy Work done ...
      • Potential(gravity) Work done ... per unit mass


24. When an alpha particle approaches a _{42}^{96}\textrm{Mo} nucleus head-on, the closest distance between the alpha particle and the nucleus is d\,. If the same alpha particle approaches a _{209}^{84}\textrm{Po} nucleus head-on, what is the closest distance between the alpha particle and the _{84}^{209}\textrm{Po} nucleus? Media:2009-1-24.jpg

  • A 0.5d\,
  • B 2.0d\,
  • C 2.2d\,
  • D 4.0d\,
  • Suggested Answer B
  • Suggested Solution
    • \begin{align} & K_{i}+\cancelto{0}{U_{i}}=\cancelto{0}{K_{f}}+U_{f}\\ & \overline{K_{i}} = \frac{Q\overline{q}}{\overline{4\pi \varepsilon_{0}}d}\\ & \therefore d \propto Q \\ & \therefore d_{2}= 2.0d \\ \end{align}
  • Suggested Explanation
    • Just looking at the quantities involved seems not enough as there is no easy formula to relate them
    • Let's try to identify the potentially useful information
      • When an alpha particle approaches a {\color{Blue}_{42}^{96}}\textrm{Mo} nucleus head-on, the closest distance between the alpha particle and the nucleus is {\color{Blue}d}. If the same alpha particle approaches a {\color{Blue}_{84}^{209}}\textrm{Po} nucleus head-on, what is the closest distance between the alpha particle and the _{84}^{209}\textrm{Po} nucleus?
    • We need to derive a relationship, but we have to first understand what is happening
      • Why is there a closest distance in the first place?
        • It will mean that it will stop. Why? Both are the alpha particle (_{2}^{4}\textrm{He} nucleus) and the nucleus are positively charged
        • So there will be electrostatic repulsive force between them, which will slow down the alpha particle until it stops at the closest distance
      • What happens when it stops?
        • F\, is NOT zero, nor is it equal for both case
        • Hmmm... something must be zero/same. velocity is zero (for both cases). But there doesn't seem to be a formula to easily use it directly.
        • What will also be zero if velocity is zero? Kinetic energy
      • What will happen to the initial kinetic energy?
        • It will change to potential energy, so lets try using conservation of energy to solve
    • K_{i}+U_{i}=K_{f}+U_{f}\,
      • Let's deal with kinetic energy first
        • Initial kinetic energy is same for both cases since it was stated same alpha particle
        • Final kinetic energy is both zero since it stops
      • Potential energy
        • Formula of potential energy of a charge in a electric field cause by a point charge/charged sphere (nucleus) \frac{Qq}{4\pi\varepsilon_{0}r}
        • Initial potential energy assuming the alpha particle is approaching from a relatively far distance (from infinity), potential energy is zero
        • Final potential energy is
          • \frac{Qq}{4\pi\varepsilon_{0}d}
          • Let Q\, to refer to charge of nucleus, q\, to refer to charge of alpha particle
      • Putting in those information, and noting down the constants and the equal quantities
        • \begin{align} & K_{i}+\cancelto{0}{U_{i}}=\cancelto{0}{K_{f}}+U_{f}\\ & \overline{K_{i}} = \frac{Q\overline{q}}{\overline{4\pi \varepsilon_{0}}d}\\ \end{align}
        • d \propto Q
        • To obtain Q\, _{{\color{Blue}42}}^{96}\textrm{Mo}, _{{\color{Blue}84}}^{209}\textrm{Po}
        • Thus, Q\, has doubled, thus d\, has doubled
        • Alternatively, \begin{align} & d \propto Q \\ & \frac{d_{2}}{d_{1}}=\frac{Q_{2}}{Q_{1}}\\ & \frac{d_{2}}{d}=\frac{84e}{42e}\\ & \therefore d_{2}= 2.0 d \\ \end{align}

  • Alternative Methods / Further investigation /Related questions
    • As a quick check, does it make sense that the distance is larger? Yes, since the electrostatic force will be bigger, so it will stop sooner and further
    • Can we use force to solve? No, as there isn't anything we can conclude about force when it stops (it is neither zero nor equal for both cases)
    • Can we use kinematic formulas such as v^{2}=u^{2}+2as\, to solve? No. Force wouldn't be constant as it approaches. Similarly, we can't use formula such as W=Fd\,
    • Can we use \Delta U =-\Delta K\, (gain in potential energy = loss of kinetic energy)? Yes, but it is much easier to use K_{i}+U_{i}=K_{f}+U_{f}\, in most cases (except when we are given potential difference)
    • Is the head on important? Well, otherwise, it will not stop momentarily but be deflected
    • Why is it that we can assume energy conserved? This is probably done in vacuum (or assumed the alpha particle didn't collide with any air molecules on the way). Anyway, it makes no sense to talk about heat loss for atoms/molecules/ions/nucleus, since temperature itself refers to the random kinetic energy of atoms/molecules/ions/nucleus.
Retrieved from "http://www.stpmwiki.com/index.php/Physics_Past_Year_Chapter_17"
Personal tools