Notes: generic problem solution steps
- 1 - Carefully read entire question. Highlight context cues that indicate which principles are involved, and information (numerical values) that will be associated with specific variables later.
- 2 - Make a visual depiction of what the problem describes, including what you are looking for.
- 3 - Identify (with variable names) information you are given, and what you are looking for. If possible, indicate it on your depiction.
- 4 - Name the physics principles which apply to the scenario described in the problem. See list of principles and equations for examples.
- 5 - List equations which are implied by the principles from step 4.
- 6 - Plot out your solution. Make any available simplifications. For example, eliminate terms which will be 0.
- 7 - Substitute values from step 3 into equations from step 6, and solve for unknowns. HANDLE UNITS CORRECTLY.
- 8 - Check that the solution has correct units. Check the sign and magnitude if possible.
2, diagrams
- If motion is involved, a motion diagram
- If forces are involved, a force diagram
4/5, principles/equations
- MUST conform to names from the list.
- Try to be complete with principles - multiple principles will apply to a single problem.
- 4 and 5 must agree - no equations without their associated principle
7, math
- Units must be handled correctly.
- Ex (good): r = sqrt((15 cm)2 + (20 cm)2) = 25 cm
- Ex (bad): r = sqrt(152 + 202) = 25 cm
- In the bad example, r being a distance must have units, but they disappeared in the middle equation, and reappeared in the right hand side (units magic).
8, checks
- State what the correct units should be.
- If you are given a range of reasonable values in lecture, state what that range was, and whether the value you calculated fell into that range.
- If you are not given a range of reasonable values, if what you are solving for is a familiar value (ex: height of a building, speed of a car, etc.) you should be able to tell if your answer is reasonable or not. 0.01 m is not reasonable for a height of a building, and 105 m/s is not reasonable for the speed of a car.
- Sign: magnitudes can not be negative, and signs of components should match the direction of the vector.
Things to remember
- You MUST organize - write out steps in order
- You may often have to go back to earlier steps to fill in details, so leave space between steps that can be filled in later.
- Simplifications = rewrite equations, omitting terms which are 0; cancel out terms which can be cancelled
- Each step requires visible work.
Example:
- One small charged ball with mass 0.93 g hangs from a string.
- String is at a 13° angle to the right of veritical.
- A plastic rod with a -2.4 nC charge is 0.86 cm to the right of the charged ball.
- The ball is in equilibrium.
- What must the charge on the ball be?
Long solution
- Step 1: Context (small charged objects, string, equilibrium)
Highlight values
- Step 2: diagram, label values (θ = 283°, r = 0.86 cm, q2 = -2.4 nC, m = 0.93 g = 0.00093 kg)
- Step 3: Looking for q1
- Step 4: Newton's 2nd, vectors (components, addition), force of gravity, coulomb's law
- Step 5: Fg = m g, Fe = ke ...
Fgx + FTx + Fex = 0, Fgy + FTy + Fey = 0.
Equations for components
- Step 6: Find components of known force (gravity).
Use y-equation to find FT.
Use x equation to find Fe.
Use Coulomb's law to find q1.
Step 7
- Fg = 0.0091 N
FTy = -0.0091 N → FT = FTy / sin(283°) = 0.0094 N
FTx = 0.0094 N × cos(283°) = 0.0021 N → Fex = -0.0021 N
Fe = 0.0021 N and |q1| = Fe r2 / (ke |q2|)
→ |q1| = 0.0021 N × (0.0086 m)2 / (|-2.4 nC| × 9.0×109 N m2/C2)
→ |q1| = 7.2 nC
Step 8
- units: Must be Coulombs (charge)
- magnitude: Expected range = 10^-10 - 10^-5. Falls in range.
- Sign: Magnitude of force equation does not tell us whether charge is + or -. Instead, we must look at whether the charges are attracting or repelling. Attracting → q1 is +.
Note
- Demonstrate units magic.
- ↑ Units magic won't be accepted for long problems.
Worksheet? Or just put on board
- 1 - 1d problem (hanging charge, find tension)
- 2 - 2d problem (incline plane)
Procedure
- Students do problem 1, steps 2-5, then interrupt to get on same page...
1d
- A small sphere hangs from a string.
- The sphere has mass of 6.4g, and charge of +25nC.
- A second sphere is 2.5cm below, with charge of +48nC.
- Find tension in string.
1d Solution
- Step 2: schematic + force diagram
- Step 3: m = 0.0064kg, q1 = 2.5×10-8C, r = 0.025m, q2 = 4.8×10-8C, looking for FT
- Step 4: Force between point charges, Newton's 2nd, equilibrium
- Step 5: ...
- Step 6: FT = m g - ke q1 q2 / r2
- Step 7: FT = 0.062N - 0.017N = 0.045N
2d
- A ball is placed on an insulating inclined plane (6.0° relative to the horizontal).
- The ball has a charge of -8.0nC, mass of 2.0g.
- A plastic rod is charged to -14nC, and held down the incline from the ball.
- How far must the rod be from the ball, if the ball is in equilibrium?
2d Solution
- Step 2: schematic + force diagram
- Step 3: θ = 6°, q1 = -8.0nC, m1 = 0.0020kg, q2 = -14nC, looking for r
- Step 4: Force between point charges, Newton's 2nd, equilibrium, incline (normal force)
- Step 5: ...
- Step 6: (only worry about x direction) FE = m g sin(θ), r = sqrt(ke q1 q2 / FE)
- Step 7: FE = 0.0020N, r = 0.022m
- Step 8: units must be in N, distance of cm is reasonable for an incline plane
Optional Example: Lab Problem
- Two balls, suspended from strings, with equal charge.
- Strings are 12cm long. Balls are 0.50g each.
- If strings make a 10° angle with the vertical, what is the charge on the balls?
Step by step solution
- Step 2: diagram
- Step 3: L = 0.12m, m = 0.00050kg, θ = 10°, looking for q
- Step 4: Force between point charges, Newton's 2nd law, vector components, equilibrium
Step 7
- Fg = m g = 0.0005kg * 9.8m/s2 = 0.0049N
- FT = Fg / cos(θ) = 0.0049N / cos(10°) = 0.0050N
- FE = FT * sin(θ) = 0.0050N * sin(10°) = 8.6×10-4
- r = 2L * sin(θ) = 0.042m
- q = sqrt(FE r2 / ke) = 1.3×10-8C
Step 8
- units = C (expected for charge).
- magnitude: 10-10 - 10-5C expected (for "large" objects)
Note
- Units magic will not be accepted on tests (give an example)
Additional Example: 2d triangle
- A +25nC charge is placed on each corner of an equilateral triangle.
- The triangle has sides of length 15cm.
- What is the magnitude of the electrical force on each charge?
Solution
- Step 2: Add force diagram for 1 charge (bottom right charge, 2 forces)
- Step 3: q1 = q2 = +25nC, r = 0.15m, θ = -60°, looking for Fnet
- Step 4: Force between 2 point charges, vector components
- Step 6: F1 = F2 = ke q2 / r2, Fnet x = F1 + F2 cos(-60°), Fnet y = F2 sin(-60°)
- Step 7: F1 = F2 = 2.5×10-4N, Fnet x = 3.8×10-4N, Fnet y = -2.2×10-4N, Fnet = 4.3×10-4N
- Step 8: force → N, magnitude of 10-4 is reasonable