# Newton's Laws

## Newton's Laws Revision

**Newton’s Laws**

There are **3** Laws of Motion which are true for **any** system.

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

**Newton’s First Law**

A body will maintain a **constant velocity** (including rest), unless an **external force** is applied to that body.

**Newton’s Second Law**

The **resultant force** on a body is **equivalent to its mass, multiplied by its acceleration**. Remember, the **acceleration** will be in the same direction as the **resultant force**. To be exact, we have the equation

F = ma

So, just as an example, in the **Resolving Forces** section, we’ve referenced the equation W = mg, **relating weight to mass and the force of gravity**. Weight is a** force**, and the value of g is given as 9.8\text{ ms}^{-2}, which is **an acceleration**.

**Newton’s Third Law**

When **two bodies collide**, they exert forces upon each other which are **equal in magnitude** but **opposite in direction**.

You might have heard the phrase, **“every action has an equal and opposite reaction”**. So, let’s say you lean against a wall. Since you are exerting a force on the wall, the wall exerts an **equal force back on you**.

**Newton’s Laws in Vector Form**

We can also apply F = ma to a **system with multiple components**, by applying the formula to **each component separately**.

**Example: **say a 4\text{ kg} particle is travelling with an acceleration \underline{a} = 2\textbf{i} + 3\textbf{j}.

Then its force F is given by F = 4(2\textbf{i} + 3\textbf{j}) = 8\textbf{i} + 12\textbf{j}.

**Example 1: Rising and Falling Objects**

A bag of mass 6\text{ kg} is picked up with a force of 90\text{ N}. Given that the only downward force is due to gravity, calculate its upward acceleration.

**[2 marks]**

First, we need to **resolve our vertical forces**. So, taking upwards to be the positive direction, we have

90 - (6 \times 9.8) = 31.2\text{ N}

Now, using Newton’s Second Law, F = ma, we have 31.2 = 6a. By extension, a = 5.2\text{ ms}^{-2}.

**Example 2: Resolving in Perpendicular Directions**

A particle is being pulled by a piece of string with a force of 17.32 \text{ N} at an angle of 30° to the horizontal. Given that there is no friction applied by the surface, and that the particle is accelerating horizontally at a rate of 3 \text{ ms}^{-2}, what is the **mass** of the particle? What is the **reaction** force R?

**[4 marks]**

Resolving horizontally: T_H = 17.32\cos 30° = 15\text{ N}

Since we have F = ma, we have 15 = 3m, so the mass of the particle is 5\text{ kg}.

Now, resolving vertically: T_V + R = W, we have 17.32\sin 30° + R = 5 \times 9.8

Rearranging for R, we have R = 49 - 17.32\sin 30° = 40.34\text{ N}.

## Newton's Laws Example Questions

**Question 1: ** A 1.5\text{ kg} ball is falling with acceleration due to gravity.

What is the force acting upon it due to gravity?

After it has hit the ground, what is the reaction force R between it and the ground?

**[2 marks]**

Using Newton’s Second Law, we have F = 1.5 \times 9.8 = 14.7\text{ N}.

Using Newton’s Third Law, we must have R = W = 14.7\text{ N}.

**Question 2: **A 5\text{ kg} particle is travelling at 4.5\text{ ms}^{-1}. If a constant force of 0.1\text{ N} is applied to oppose its motion, how long does it take to stop?

**[2 marks]**

F = ma gives a = \dfrac{-0.1}{5} = -0.02\text{ ms}^{-2}

From here,

v = u + at gives 0 = 4.5 - 0.02t

Then, after rearranging,

t = \dfrac{4.5}{0.02} = 225\text{ s}

**Question 3: **A particle is decelerating, such that \underline{a} = \begin{pmatrix}-9\\ -3\end{pmatrix}. Given that its mass is 0.5\text{ kg}, what force is being applied to the particle?

**[1 mark]**

\underline{F} = m\underline{a}

Then,

\underline{F} = 0.5\begin{pmatrix}-9\\ -3\end{pmatrix} = \begin{pmatrix}-4.5\\ -1.5\end{pmatrix}

## Newton's Laws Worksheet and Example Questions

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