Attach three strings to a single pin or knot at a central point O on the board.
Hang known weights from two of the strings β these represent forces Fβ and Fβ at different angles.
The third string holds a counterbalancing weight (or is pulled by hand so the knot stays at O).
Mark the point O and trace each string direction onto the paper with a pencil.
Record the magnitude of each force (Fβ, Fβ) against their respective string lines.
Remove the paper. Draw vectors to scale along the traced lines.
Complete the parallelogram: draw lines parallel to each vector from the tip of the other.
The diagonal of the parallelogram from O gives the resultant R β measure its length and angle.
The diagonal of the parallelogram represents the resultant force in both magnitude and direction.
The resultant balances the equal and opposite equilibrant force (the third string).
Measured resultant magnitude and direction match the calculated values.
Think of two people pulling a boat at different angles with ropes β the boat doesn't go in either person's direction but somewhere in between, exactly as predicted by the parallelogram.
State the scale you used. Show the parallelogram construction. State: 'Resultant = diagonal of parallelogram from the point of application.'
2Conservation of Momentum
Linear air track (or smooth runway with two trolleys)
Two trolleys of known mass with Velcro/sticky bumper (for perfectly inelastic collision)
Ticker-tape timer and tape (or two light gates with data-logger)
Metre rule
Balance to measure trolley masses
Set up the air track horizontally. Measure and record masses mβ and mβ.
Attach ticker tape to trolley A. Thread it through the ticker-tape timer.
Hold trolley B stationary at the centre. Start the ticker-tape timer.
Release trolley A from the end β it travels and collides with trolley B, sticking together.
From the ticker tape, measure the dot-spacing before collision (for uβ) and after (for v).
Calculate momentum before: p_before = mβuβ (trolley B has uβ = 0)
Momentum before collision (mβuβ) β momentum after collision ((mβ+mβ)v).
Any small difference is due to friction β the track minimises this.
The total momentum is conserved within experimental error.
Like a train carriage rolling into a stationary one and coupling β the joined carriages move on more slowly but with the same total momentum as the original moving carriage.
Must state 'system is closed' (no external forces). Show all calculations. Write: 'Within experimental error, momentum is conserved.'
3Demonstration of Atmospheric Pressure
Empty aluminium soft-drink can (or round tin)
Bunsen burner / hotplate
Tongs
Bowl of cold water
Alternative: Magdeburg hemispheres and vacuum pump
Place a small amount of water (~10 mL) in the bottom of the can.
Heat the can on the Bunsen burner until the water boils vigorously and steam pours from the opening β this drives most of the air out, replacing it with steam.
Using tongs, quickly invert the can and plunge its opening into a bowl of cold water.
Observe what happens.
Magdeburg hemispheres alternative: join the two hemispheres, pump out the air with the vacuum pump; two students try to pull them apart.
The can is instantly crushed inward with a loud bang.
The steam condenses rapidly, creating very low pressure inside the can.
Atmospheric pressure (~101 kPa) outside pushes in from all sides, collapsing the can.
For Magdeburg hemispheres: the joined hemispheres cannot be pulled apart because atmospheric pressure holds them together β until air is let back in.
The can is like a bag with air pumped out β the outside pressure, normally balanced by internal pressure, now has nothing to push against and crushes inward.
State clearly: 'The crushing force is atmospheric pressure (~101 kPa) acting on the external surface when internal pressure is reduced.'
4Verify Archimedes' Principle
Overflow can (Eureka can) filled to the spout with water
Stone or metal object of known weight
Spring balance (0β5 N range)
Measuring cylinder or small beaker to collect displaced water
Ruler, string
Weigh the stone in air using the spring balance. Record weight Wβ.
Fill the overflow can to the level of the spout. Place collection beaker under the spout.
Lower the stone gently into the water using the string β displaced water flows into the beaker.
Read the new spring balance reading Wβ (apparent weight in water).
Weigh the collected overflow water using the spring balance. Record Wβ.
Calculate upthrust = Wβ β Wβ. Compare with Wβ.
Upthrust (Wβ β Wβ) equals the weight of water displaced (Wβ) within experimental error.
This verifies Archimedes' Principle: upthrust = weight of fluid displaced.
When you step into a bath and water spills over the edge, the weight of that spilled water equals the upward push you feel β that's exactly what this experiment shows.
Show calculation: Upthrust = Wβ β Wβ. Weight displaced = Wβ. Conclusion: upthrust = weight of displaced fluid. State units (N) throughout.
5Demonstration of an Energy Conversion
Simple pendulum: string (~1 m), heavy bob (50β100 g), retort stand and clamp
Alternatively: trolley on ramp, metre rule, ticker tape
Ruler to measure height of bob
Set up the pendulum on the retort stand. Mark the equilibrium position of the bob with chalk.
Pull the bob to one side (height h above equilibrium) β this is the starting position.
Release from rest. The bob swings through the equilibrium point and rises to an equal height on the other side.
Using a ruler, measure the height at both extreme positions and at the centre (lowest point).
Discuss: at top of swing, all energy is GPE = mgh. At bottom, all energy is KE = Β½mvΒ².
The bob rises to approximately the same height on each side β demonstrating conservation of energy.
At the extreme positions, the bob is momentarily stationary (KE = 0, GPE = maximum).
At the lowest point, the bob moves fastest (KE = maximum, GPE = minimum/zero).
Slight loss of height over successive swings is due to air resistance (energy converted to heat).
A skateboarder on a half-pipe converts PE at the top of each side into KE at the bottom and back again β the skater reaches the same height each time (ignoring friction).
Write: 'Total mechanical energy is conserved (PE + KE = constant), with small losses to friction/air resistance.' Quote the Principle of Conservation of Energy.
6Demonstration of Circular Motion
Rubber bung (~50 g)
Glass tube (~30 cm, fire-polished ends)
String (1β1.5 m) threaded through the tube
Paper clip and small hanging mass (e.g. 50 g weight) attached below the tube
Metre rule, stopwatch
Thread the string through the glass tube. Tie the rubber bung to one end; attach a known hanging mass (mg) to the lower end via a paper clip.
Holding the tube vertically, swing the bung in a horizontal circle above the tube.
Adjust the speed of rotation until the paper clip just remains stationary (the string neither rises nor falls).
At this point, the tension T in the string = mg (weight of hanging mass) = centripetal force.
Measure the radius r of the circular path and use a stopwatch to time 20 revolutions to find period T.
Verify: mg = mvΒ²/r = mΟΒ²r.
When the bung is spun at the correct speed, the hanging mass remains stationary β indicating the centripetal force equals mg.
If the bung is spun faster, the paper clip rises (less string below = smaller radius).
If slower, the paper clip falls.
This confirms: centripetal force = mvΒ²/r.
Like a ball on a string swung around your head β the string tension (which you feel in your hand) is the centripetal force pulling the ball inward, exactly like gravity keeping a satellite in orbit.
State clearly which force provides the centripetal force (tension in string). Formula: F = mvΒ²/r. Note: centripetal force is not a separate force β it is the name for the net inward force.
7Demonstration of Simple Harmonic Motion
Spring (Hooke's law spring, spring constant k known or to be found)
Known mass m (e.g. 100β200 g)
Retort stand and clamp
Ruler (clamped vertically) and pointer
Stopwatch
Alternatively: simple pendulum
Clamp the spring to the retort stand. Attach mass m to the lower end.
Allow the springβmass system to reach equilibrium. Mark the equilibrium position on the ruler.
Pull the mass down by a small distance x (amplitude ~2β3 cm) and release.
The mass oscillates up and down. Time 20 complete oscillations with a stopwatch.
Calculate period T = total time / 20.
Repeat with different amplitudes β observe that T remains the same.
Note: the acceleration is always directed back towards the equilibrium position, and is proportional to displacement.
The mass oscillates about the equilibrium position with a constant period.
The period is independent of amplitude (as long as the elastic limit is not exceeded).
This satisfies the two conditions for SHM: a β x and a is directed towards equilibrium.
Period formula: T = 2Οβ(m/k) β doubling the mass increases T by β2.
Like a bouncing car suspension β the heavier the car the slower it bounces, but it always bounces at the same rate regardless of how far you push it down (small displacements).
Must state BOTH conditions for SHM: (1) a β displacement, (2) a is always directed towards the equilibrium position. The period is independent of amplitude.