01 · Shared acceleration
The pulling force on B is instead 24 N right, with the same masses and resistances. Find acceleration.
Hint
Use the total mass and both resistances.
Worked solution
24 − 3 − 5 = 8a gives a = 2 m/s² right.
Understand · explore · practise
Connected particles on a horizontal surface
Find shared acceleration from the whole system and tension from an individual body, including resistance, unknown forces and changing the pulled end.
Before you startNewton’s second and third laws, light strings and constant acceleration.
01 / Use the connection to relate the accelerations
The force diagrams remain separate.
A taut inextensible straight string keeps the separation fixed.
In this horizontal arrangement both particles have the same signed acceleration. A light string transmits equal tension magnitudes to its ends. Tension pulls each particle toward the other. The string cannot push and the shared-acceleration condition ends if it becomes slack.
Whole system, then one particleExplore
A (3 kg, left) and B (5 kg, right) are joined by a taut light inextensible string on a horizontal surface. Resistances of 3 N on A and 5 N on B oppose the stated motion. Pull B right or A left. The straight taut connection gives both particles the same acceleration.
02 / Find acceleration from the whole system
Internal tension cancels when both particles are included.
A has mass 3 kg and resistance 3 N; B has mass 5 kg and resistance 5 N. B is pulled right by 40 N while both move right.Worked example
40 − 3 − 5 = (3 + 5)a
Only external horizontal forces remain in the combined equation.
a = 4 m/s² right
Both particles share this acceleration.
Vertical forces balance separately
The horizontal pull adds no vertical component in this model.
Watch: internal tension cancels in the combined equation
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Smooth surface
Two connected masses 2 kg and 6 kg are on a smooth horizontal surface, with an external 32 N pull on the 6 kg particle. Find shared acceleration.
Hint
No friction or other horizontal resistance is stated.
Worked solution
32 = 8a, so a = 4 m/s² in the pulling direction.
03 / Return to one particle to find the tension
Internal cancellation removes T from the system equation.
Return to the 40 N worked example.Worked example
For A: T − 3 = 3a
The 40 N pull acts on B, not on A.
T − 3 = 12, so T = 15 N
Use the shared acceleration 4 m/s².
For B: 40 − T − 5 = 5a
A separate check gives 40 − 15 − 5 = 20.
03 · Tension at 24 N pull
Find T in question 1.
Hint
Use the 3 kg particle.
Worked solution
T − 3 = 3 × 2, so T = 9 N.
04 · Smooth tension
Find string tension in question 2.
Hint
The unpulled 2 kg mass has only tension horizontally.
Worked solution
T = 2 × 4 = 8 N. This is smaller than the external pull of 32 N.
05 · Why not total force?
Why is T not 40 N in the worked example?
Hint
The external pull also accelerates B and overcomes its resistance.
Worked solution
B’s equation is 40 − T − 5 = 5a. T is the force transmitted to A; it is not the entire external force.
04 / Changing the pulled end can change tension
Total external force may stay the same while the internal force changes.
06 · Pull the other end
Now pull A left by 40 N while both move left. The resistance magnitudes remain 3 N and 5 N. Find acceleration magnitude.
Hint
Take left positive and use the whole system.
Worked solution
40 − 3 − 5 = 8a gives 4 m/s² left.
07 · New tension
Find T for question 6.
Hint
B is now the unpulled particle.
Worked solution
For B, T − 5 = 5 × 4, giving T = 25 N. The same total acceleration magnitude does not imply the same tension.
08 · Compare diagrams
When the pulled end changes, does tension on A switch from rightward to leftward?
Hint
A remains to the left of B and the string pulls toward B.
Worked solution
No. The tension on A points right and on B points left in both arrangements. The external pull and resistance directions change, while the string’s geometric pull directions stay the same.
05 / Use an individual equation to recover an unknown
A known tension can supply the acceleration.
09 · Unknown pull
In the original rightward arrangement, T = 12 N. Find the external pull on B.
Hint
First use A’s equation.
Worked solution
12 − 3 = 3a gives a = 3 m/s². Whole system: P − 8 = 8 × 3, so P = 32 N.
10 · Unknown mass
On a smooth horizontal surface, an unpulled particle has unknown mass m and is linked to a 6 kg particle pulled by 30 N. Tension is 12 N. Find a and m.
Hint
Start with the pulled particle because its mass is known.
Worked solution
30 − 12 = 6a gives a = 3 m/s². Then 12 = ma gives m = 4 kg.
11 · Unknown resistance
A 2 kg unpulled particle has resistance R and is linked to a 4 kg particle with resistance 4 N. An external 24 N pull produces acceleration 2 m/s². Find R and tension.
Hint
Use the whole system, then the unpulled particle.
Worked solution
24 − R − 4 = 6 × 2 gives R = 8 N. For the 2 kg particle, T − 8 = 2 × 2, so T = 12 N.
06 / Use the shared acceleration for each particle’s motion
Constant force assumptions justify a constant-acceleration stage.
12 · Speed and travel
In the 40 N worked example, both start from rest. Find speed and distance after 3 s.
Hint
Use a = 4 m/s².
Worked solution
v = 4 × 3 = 12 m/s; s = ½ × 4 × 3² = 18 m for each particle.
13 · Time to a speed
For the same model, how long does it take to increase speed from 2 to 10 m/s?
Hint
Use v = u + at.
Worked solution
10 = 2 + 4t gives t = 2 s.
07 / Check that the string model remains possible
A negative required tension is a warning about the assumed connection.
14 · A negative tension result
A proposed model of two bodies joined only by a string gives T = −6 N. Can this be interpreted as a 6 N push by the string?
Hint
A string transmits a pull, not compression.
Worked solution
No. The assumed taut-string model is inconsistent with that result; the string may become slack and the shared-acceleration equations need reconsidering. A rigid rod is a different model.
Keep resistance opposite the actual or stated motion, not automatically opposite acceleration. Never include an internal tension twice in the whole-system resultant.
08 / Combine bodies for acceleration, separate them for tension
Check signs and the connection assumptions.
Draw each body’s forces, relate accelerations using the taut connection, add compatible equations to remove internal tension, then use one body to recover it. Check both equations and whether the string can sustain the calculated force.
Section 1 of 8 · Use the connection to relate the accelerations