Physics Problem Solving

This site is to help you solve physics problems. It is only helpful if you follow the steps.
Sample problems will be added throughout the semester, as each unit is taught. These will be links to another page. Just look for

Steps to follow

  1. Read the problem carefully. Hopefully the probelm makes sense and has a ring of familiarity. (This should be the case in all test or exam questions.) If not, to be honest with you so what! Just get on with step two.
  2. Write down data as it is read in the problem; if you have no idea how to solve the problem do not worry about it. The analysis will take care of this for you. Often, it is very difficult to solve the problem in your head, so don't try. Let the problem solving steps take care of this.
    Its like hopping on a sailboard in 25K winds, if you ever started thinking of what you about to do you'd probably never sail the board.
  3. Write each bit of information down with the correct symbol, value and units.
    For example if the problem says "the initial velocity is 4.56 m/s" then you write this down as
    v1 = 4.56 m/s
    and then go on to the next bit of information.
    Watch you subscripts. For example, will always infer the angle of incidence.
    Continue doing this until all the given information is recorded in this format.
  4. At the same time make a diagram or sketch of what is happening if the problem. Especially true in ray diagrams & thin lens problems. Sometimes the sketch is the answer or the answer is derived or measured from the diagram as in a vector problem. Use rules & pencil for accurate drawing. Be aware of using a scale to ensure some accuracy in the way the diagram looks.
  5. Somewhere you will encounter the "QUESTION". Do as above only this time an unknow will be used instead of a number. And again don't forget the units of this unknown.
    For example: the problem say to determine the heat capacity of a block of zinc. You write down
    C = ? J/goC (Note g could be kg)
  6. The problems now generally take two routes;
    1. Using a formula
    2. A pattern to follow
    You must decide which of the two will solve the problem, but this should be very obvious.
  7. If a formula is to be used you now substitute all values including the known into the formula. Substitute directly, do not do any manipulations at all.
  8. Now you can do the algebra on the substituted formula
  9. If this solved value is the answer to the question then make a statement which answers the question asked in the problem. And don't forget the units.

For a more in detailed look at problem solving try
Solving Physics Problems; a little long but hey, if you are having trouble ......

Examples

Unit I -- Light and Optics

Unit II: Energy

A Couple of Heat Problems

  1. A block of iron alloy has a specific heat capacity of 456.2 J/goC. If 135 grams of this alloy is heated from 17.8oC to 32.5oC, calculate the amount of heat in kilojoules is absorbed by this metal alloy.

    SOLUTION
    Mass of metal m = 135 g
    Specific heat capacity c = 456.2 J/goC
    T1 = 17.8oC
    T2 = 32.5oC
    Quantity of heat Q = ? kJ/goC

    The first step is to releaze that a t must be found. Subtract the two temperatures and
    t = 14.7 oC
    Now the formula Q = mtc
    Substitute into the formula, directly with no math Q = 135 x 14.7 x 456.2 (unit left out)
    Do the arithmetic and Q = 905328.9 J
    Read the question; this is not the answer & it must also be rounded off correctly. Three signifigant digits are required.
    The amount of heat absorbed is 905 kJ. or 9.05 x 102 kJ

  2. Calculate the specific heat capacity of copper when a 167.2 gram sample is given 5.03 x 102 kJ of heat energy. The temperature of the copper changes from 22.4oC to 33.7oC.

    SOLUTION
    A slightly shorter but exceptable format
    c = ? kJ/goC
    m = 167.2 g
    Q = 5.03 x 102 kJ
    = 33.7oC - 22.4oC = 11.3 oC

    The formula is Q = mtc
    Substitute into the formula, directly with no math.

    5.03 x 102 kJ = 167.2 g x 11.3 oC x ? kJ/goC
    This can now be solved. Answer is 0.266227
    The specific heat capacity of copper is 0.266 kJ/goC.
    If you cannot get this value then bring to class for direct, personal assistance.

  3. Calculate the final tmperature when 45.6 grams of water at 17.8oC is mixed with 67.4 grams of warm water at 52.4oC. You can assume no heat lost to the calorimeter.
There are two cups of water; one hot and one cold. The hot water will lose heat to the cold water and applying the princilpe of heat eschange each will attain the same temperature at equilibruim.

Cold Water       m = 45.6 g        t1 = 17.8oC        t2 = ?o        c = 4.18 J/goC

Hot Water       m = 67.4 g        t1 = 52.4oC        t2 = ?o        c = 4.18 J/goC

Since heat gained equals heat lost Qhot = Qcold and using the formula is Q = mtc
the following equation is written

45.6 x (t2 - 17.8) x 4.18 = 67.4 x (52.4 - t2) x 4.18

And you solve this for t2

Note: the unknown t2 is in a different position on each side of the equation. Instead of using a vector heat flow principle I used common sence where cold water becomes warmer and hot water becomes colder. In either case there is a switch in the sign of t2.

You should have got 38.4oC

If you didn't then get some math help.