# Connected Particles

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Connected Particles

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.

A Level

## 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 $t = 0.43\text{ seconds}$

A Level

## 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

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

A Level

## 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

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

We can conclude that weight $\text{B}$ rises.

A Level

## Example Questions

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

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

$T = 33.32\text{ N}$

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

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

A Level

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