01 · Another pair
A is 2 kg and B is 3 kg. Find acceleration magnitude and tension after release.
Hint
Add 29.4 − T = 3a and T − 19.6 = 2a.
Worked solution
a = 9.8/5 = 1.96 m/s²; T = 19.6 + 2 × 1.96 = 23.52 N. B descends.
Understand · explore · practise
Two hanging particles over a pulley
Solve two hanging-particle equations for acceleration and tension using compatible coordinates, find unknown masses and restrict motion to the taut-string stage.
Before you startNewton’s second law, weight, light strings and constant acceleration.
01 / Choose a positive direction separately for each particle
The accelerations have equal magnitudes and opposite vertical directions.
Use one scalar a for compatible motion along the string.
If B descends, take downward positive for B and upward positive for A. A taut inextensible string over a fixed pulley gives equal displacement, speed and acceleration magnitudes along these coordinates. These are not equal acceleration vectors. A light string over a smooth pulley has the same tension magnitude on both sides.
Two masses, one taut stringExplore
Particles A and B hang on opposite sides of a fixed smooth pulley. The connecting string is light, taut and inextensible. They are released from rest with both clear of the ground. Set g = 9.8 m/s². Select masses; the heavier particle descends and the lighter rises. Equal masses stay at rest in this ideal model.
02 / Write one equation for each particle
Weight and tension belong to each body’s own diagram.
A has mass 3 kg and B has mass 5 kg. Both are released from rest, with B descending.Worked example
For B: 49 − T = 5a
Downward positive for B.
For A: T − 29.4 = 3a
Upward positive for A.
Add: 19.6 = 8a
The two chosen scalar directions are compatible with the connection.
a = 2.45 m/s²; T = 36.75 N
Use T = 29.4 + 3a, then check the B equation.
Watch: opposite axes give compatible scalar equations
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Check the heavier body
Check the 5 kg body in the worked example.
Hint
Its net downward force should equal 5a.
Worked solution
49 − 36.75 = 12.25 = 5 × 2.45 N.
03 / Derive the general result from the two equations
Use positive masses and keep the direction convention explicit.
For B of mass M heavier than A of mass m: a = (M − m)g/(M + m).
From Mg − T = Ma and T − mg = ma, addition gives (M − m)g = (M + m)a. Substitution gives T = 2Mmg/(M + m). These formulas assume the same fixed smooth pulley and taut light string model.
03 · Direction from masses
Let A have mass 6 kg and B have mass 2 kg. Which particle descends, and what are a and T?
Hint
Choose A downward and B upward since A is heavier.
Worked solution
A descends with a = (6 − 2)9.8/8 = 4.9 m/s². T = 2 × 6 × 2 × 9.8/8 = 29.4 N.
04 · Equal masses
Both masses are 4 kg and are released from rest. Find a and T.
Hint
The weight difference is zero.
Worked solution
a = 0 and T = 39.2 N. With zero initial velocity they remain at rest while the ideal assumptions hold.
05 · Bounds
Explain why the acceleration magnitude is less than g when both masses are positive.
Hint
Compare |M − m| with M + m.
Worked solution
For positive masses, |M − m| < M + m, so |a| < g. The string reduces the descending particle’s acceleration below free fall.
04 / Use a measured acceleration to find a mass ratio
The two masses affect both the driving difference and total inertia.
06 · A ratio
B is heavier and acceleration magnitude is g/4. Find B’s mass divided by A’s mass.
Hint
Set (M − m)/(M + m) = 1/4.
Worked solution
4M − 4m = M + m gives 3M = 5m, so M/m = 5/3.
07 · Unknown mass
A is 4 kg; B is heavier. The acceleration is 1.96 m/s². Find B’s mass.
Hint
Here a/g = 1/5.
Worked solution
(M − 4)/(M + 4) = 1/5 gives 5M − 20 = M + 4, hence M = 6 kg.
08 · Tension evidence
A is 2 kg and rises with tension 24 N. Find acceleration and the heavier mass B.
Hint
First use T − mg = ma for A.
Worked solution
a = (24 − 19.6)/2 = 2.2 m/s². For B: M(9.8 − 2.2) = 24, so M = 60/19 kg ≈ 3.16 kg.
05 / Use SUVAT only before a new event changes the system
Both particles travel the same distance while the connection stays taut.
09 · First second
For the 3 kg and 5 kg worked example, find each speed and distance moved after 1 s from rest.
Hint
Use a = 2.45 m/s².
Worked solution
Speed = 2.45 m/s; distance = ½ × 2.45 = 1.225 m. The directions are opposite.
10 · Ground event
The heavier body starts 2 m above the ground in that example. Find the time and speed just before it reaches the ground.
Hint
Use distance 2 m with a = 2.45.
Worked solution
2 = ½ × 2.45t² gives t = √(4/2.45) ≈ 1.28 s. Speed = √(2 × 2.45 × 2) = √9.8 ≈ 3.13 m/s. This is the end of the original stage.
11 · Can the model reach 2 seconds?
Can the unchanged acceleration be used to predict motion at t = 2 s in question 10?
Hint
Compare 2 s with the ground-event time.
Worked solution
No. The heavier body reaches the ground at about 1.28 s. Contact and possible slack change the forces; a new stage must be modelled.
06 / A heavier body can initially be moving upward
Acceleration follows the forces; velocity follows the initial condition too.
12 · Initial upward velocity
In the 3 kg and 5 kg example, the heavier body initially moves upward at 4.9 m/s, with the string taut and enough clearance. How long until it reverses?
Hint
For the heavier body, choose downward positive: u = −4.9 and a = 2.45.
Worked solution
0 = −4.9 + 2.45t gives t = 2 s. It first slows while rising, then descends if no other event intervenes.
13 · Initial rise
How far does the heavier body rise before reversing in question 12?
Hint
Use v² = u² + 2as with downward positive.
Worked solution
0 = 4.9² + 2 × 2.45s gives s = −4.9 m: a rise of 4.9 m. The lighter body descends 4.9 m during this stage.
07 / Keep vector and scalar statements distinct
A convenient sum of equations is not a single vertical particle equation.
14 · A tempting total
Why is (5 − 3)g = (5 + 3)a valid here even though the masses accelerate in opposite vertical directions?
Hint
The equations were written along separate compatible motion coordinates.
Worked solution
It is the sum of two scalar equations with downward positive for the heavier mass and upward positive for the lighter. It is not the vertical vector equation of a 8 kg particle. In a common upward vector axis, the two acceleration terms have opposite signs.
A rough pulley may produce unequal tensions; a string with appreciable mass changes the light-string model. Do not apply the formulas unchanged if these assumptions are removed.
08 / Draw two diagrams and choose compatible axes
Check tension with both bodies.
State the string and pulley assumptions, use the heavier body’s downward direction and lighter body’s upward direction, solve the paired equations and then calculate motion. Stop at the first event that changes the connection or forces.
Section 1 of 8 · Choose a positive direction separately for each particle