# Connected Particles

## Connected Particles Revision

**Connected Particl**es

By a **connection**, we mean a system where **multiple particles are linked by another force**. For example, a car pulling a caravan is **connected** by the coupling, which provides **tension** on both objects.

Make sure you are happy with the following topics before continuing.

**Assumptions**

First off, we have to assume that **pulleys, pegs** and other potential external factors are **smooth**, unless explicitly told otherwise.

We’ll also treat multiple **connected** particles as **one mass**, with the **same velocity** **and acceleration**. Without this, we could assume the **connection** **has failed** between the two particles, i.e. a **linked** piece of string has snapped, or slackened.

We’re also going to **assume that Newton’s Laws hold**, especially the **Second Law**, F = ma. In fact, we’re going to use F = ma in the direction that each particle moves individually.

Otherwise, we **resolve forces** exactly how we would anyway, taking each particle separately.

**Example 1: Yo-yo System**

Back in Question 1 in the Forces section, we had two balls **attached by taut string**, **suspended by a smooth peg**.

We said that Ball 2 has twice the mass of Ball 1. Given that Ball 1 has a mass of 60\text{ g} and starts 30\text{ cm} below the peg, calculate the tension in the string and the time taken for Ball 1 to reach the peg.

**[4 marks]**

Resolving for Ball 1:

T - (0.06 \times 9.8) = T - 0.588 = 0.06a

Resolving for Ball 2:

(0.12 \times 9.8) - T = 1.176 - T = 0.12a

From here, we have

0.12a = 2T - 1.176 = 1.176 - T

which can be rearranged to

3T = 2.352, or T = 0.784\text{ N}

Substituting this back into one of our equations, we have 1.176 - 0.784 = 0.12a, meaning that Ball 2 has a downward acceleration of a = 3.27\text{ ms}^{-2}. This means that Ball 1 accelerates towards the peg at 3.27\text{ ms}^{-2} \text{ (to }2\text{ dp)}.

Using SUVAT, s = ut + \dfrac{1}{2}at^2 gives

0.3 = \dfrac{1}{2} \times 3.27 \times t^2, so t = 0.43\text{ seconds}

**Example 2: Rough Planes**

Now, imagine a system of two weights attached by a **thin piece of taut rope**, both of mass 4\text{ kg}. Weight \text{A} is rested on a rough table with a coefficient of 0.2, and weight \text{B} is suspended below a **smooth pulley**. What is the acceleration of both weights? What is the tension in the rope?

**[5 marks]**

Weight \text{A} (vertical):

R = mg = 4 \times 9.8 = 39.2\text{ N}

Weight \text{A} (horizontal):

T - \mu R = ma, giving T - (0.2 \times 39.2) = T - 7.84 = 4a

Weight \text{B} (vertical):

mg - T = ma, giving 39.2 - T = 4a

Therefore,

4a = 39.2 - T = T - 7.84

which can be rearranged to

2T = 47.04, or T = 23.52\text{ N}

Substituting back into one of our equations, we have 23.52 - 7.84 = 4a

giving

a = \dfrac{23.52 - 7.84}{4} = 3.92\text{ ms}^{-2}

**Example 3: Rough Inclined Planes**

Here’s another system of two weights attached by a** thin piece of taut rope**. Weight \text{A}, of mass 9\text{ kg}, is now on an inclined plane at an angle of 45° with a coefficient of 0.1, and weight \text{B}, of mass 3\text{ kg} is suspended below a **smooth pulley**. Does weight \text{B} rise or fall? What is the rate of acceleration of both weights? What is the tension in the rope?

**[7 marks]**

Weight \text{A} (perpendicular to motion):

R = mg\cos 45° = (9 \times 9.8)\cos 45° = \dfrac{88.2}{\sqrt{2}}

Weight \text{A} (parallel to motion, down slope):

mg\sin 45° - T - F = mg\sin 45° - T - \mu R = ma

giving

(9 \times 9.8)\sin 45° - T - \left( 0.1 \times \dfrac{88.2}{\sqrt{2}}\right)

= \dfrac{88.2}{\sqrt{2}} - T - \dfrac{8.82}{\sqrt{2}} = 9a.

Weight \text{B} (upwards):

T - mg = ma, giving T - (3 \times 9.8) = 3a

or

T - 29.4 = 3a

Equating the two equations:

9a = 3T - 88.2 = \dfrac{79.38}{\sqrt{2}} - T

So

4T = \dfrac{79.38}{\sqrt{2}} + 88.2 = 144.33

giving

T = 36.08\text{ N(to }2\text{ dp)}

Therefore, we have

9a = (3 \times 36.08) - 88.2

so

a = 2.23\text{ ms}^{-2}

## Connected Particles Example Questions

**Question 1: **We have two balls attached by taut string, suspended by a smooth peg.

If Ball \text{A} has mass 100\text{ g} and Ball \text{B} has mass 80\text{ g}, calculate the tension in the string and the acceleration of Ball \text{A}.

**[4 marks]**

**Note: **Here, “0.1g” is not the same as 0.1\text{ grams}, it is 0.1\text{ kg} by the acceleration due to gravity, g. The same rule applies for 0.08g.

Resolving Ball \text{A}:

(0.1 \times 9.8) - T = 0.1a

Resolving Ball \text{B}:

T - (0.08 \times 9.8) = 0.08a

Therefore, we have

a = 9.8 - 10T = 12.5T - 9.8

So

22.5T = 19.6

giving

T = 0.871\text{ N (to }3\text{ dp)}

By extension, a = 1.088\text{ ms}^{-2}\text{ (to } 3\text{ dp)}

**Question 2:** For two bricks of mass 4\text{ kg} linked by a rope on a smooth pulley, find the tension in the rope. Assume that the surface that brick \text{A} has a coefficient of friction of 0.7.

**[3 marks]**

Brick \text{A} (vertical):

R = 4 \times 9.8 = 39.2\text{ N}

Brick \text{A} (horizontal):

T - \mu R = ma

gives

T - (0.7 \times 39.2) = 4a

Brick \text{B} (vertical):

(4 \times 9.8) - T = 4a

So

T - 27.44 = 39.2 - T

meaning

T = 33.32\text{ N}

**Question 3: **A truck on a flat surface is towing a car up a hill with an incline of 30°. Given that the truck weighs 2000\text{ kg}, and the car weighs 1000\text{ kg} and is exerting a thrust of 2500 \text{ N}, what thrust is required by the truck to pull the car up the hill with an acceleration of 1\text{ ms}^{-2}?

Assume the coefficient of friction of both the hill and the flat surface are 0.25.

**[7 marks]**

Resolving truck (vertically):

R = 2000 \times 9.8 = 19600\text{ N}

Resolving truck (horizontally):

\text{Thrust} - T - (0.25 \times 19600) = 2000

Resolving car (perpendicular to F):

(1000 \times 9.8)\cos 30° = 8487 = R

Resolving car (parallel to F):

2500 + T - (0.25 \times 8487) - (1000 \times 9.8)\sin 30° = 1000

From the last equation, we have

T = 1000 + 4900 + 2122 - 2500 = 5522\text{ N}

Back to the truck:

\text{ Thrust} - 5522 - 4900 = 2000

which gives

\text{Thrust} = 12422\text{ N}

## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.