From Mechanics

Theory

Exercises

Key Points

Momentum is defined as the product of the mass and velocity of a body.

Momentum = \displaystyle mv, or in vector form, Momentum = \displaystyle m\mathbf{v}

If the velocity of a body, of mass \displaystyle m , changes from \displaystyle u to \displaystyle v , then

Change in momentum or impulse = \displaystyle mv - \displaystyle mu, or in vector form, \displaystyle m\mathbf{v}-m\mathbf{u}

We often use

\displaystyle I=mv-mu , or in vector form, \displaystyle \mathbf{I}=m\mathbf{v}-m\mathbf{u}

The relationships \displaystyle I=Ft, or in vector form, \displaystyle \mathbf{I}=\mathbf{F}t can be used where \displaystyle F, or in vector form, \displaystyle \mathbf{F} is constant or may be assumed to be constant and represents the average force.

Example 15.1

Find the momentum of a car, of mass 1.2 tonnes, travelling in a straight line at 30 \displaystyle \text{m}{{\text{s}}^{-1}}.

Solution

Note that 1.2 tonnes is 1200 kg. Using Momentum = mv gives:

\displaystyle \begin{align} & \text{Momentum}=1200\times 30 \\ & =36000\text{ Ns} \end{align}

Example 15.2

A car, of mass 1.1 tonnes, which was initially travelling at 6 \displaystyle \text{m}{{\text{s}}^{-1}} is brought to rest in 2.2 seconds by a wall. Find the average force exerted by the wall on the car.

Solution

First find the impulse (change in momentum).

\displaystyle \begin{align} & \text{Impulse }=mv-mu \\ & =1100\times 0-1100\times 6 \\ & =-6600\text{ Ns} \end{align}

Use the formula \displaystyle \text{Impulse }=Ft

\displaystyle \begin{align} & -6600=F\times 2\textrm{.}2 \\ & F=-3000\text{ N} \end{align}

The average force has magnitude 3000 N

Example 15.3

The speed of a car, travelling in a straight line, is reduced from 20 \displaystyle \text{m}{{\text{s}}^{-1}} to 15 \displaystyle \text{m}{{\text{s}}^{-1}} in 20 seconds. The mass of the car is 1200 kg. Find the average force acting on the car.

Solution

First find the impulse (change in momentum).

\displaystyle \begin{align} & \text{Impulse }=mv-mu \\ & 1200\times 15-1200\times 20 \\ & =-6000\text{ Ns} \end{align}

Use the formula \displaystyle \text{Impulse }=Ft

\displaystyle \begin{align} & -6000=F\times 20 \\ & F=-300\text{ N} \\ \end{align}

Example 15.4

A ball has a mass of 200 grams. It initially travels horizontally at 10 \displaystyle \text{m}{{\text{s}}^{-1}}. After being hit it travels at 16 \displaystyle \text{m}{{\text{s}}^{-1}} at an angle of 60\displaystyle {}^\circ above the horizontal. Find the initial and final momentum of the ball and the change in momentum.

Solution

The diagrams show the initial and final velocities of the ball and the unit vectors \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.

\displaystyle \begin{align} & \text{Initial Momentum }=\text{ m}\mathbf{u} \\ & =0\textrm{.}2\times 10\mathbf{i} \\ & =2\mathbf{i} \text{ Ns} \end{align}

\displaystyle \begin{align} & \text{Final Momentum }=m\mathbf{v} \\ & =0\textrm{.}2(16\cos 60{}^\circ \mathbf{i}+16\sin 60{}^\circ \mathbf{j}) \\ & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j} \text{ Ns} \end{align}

\displaystyle \begin{align} & \text{Change of Momentum }=m\mathbf{v}-m\mathbf{u} \\ & =3\textrm{.}2\cos 60{}^\circ \mathbf{i}+3\textrm{.}2\sin 60{}^\circ \mathbf{j}-2\mathbf{i} \\ & =-0\textrm{.}4\mathbf{i}+2\textrm{.}77\mathbf{j} \text{ Ns} \end{align}

Example 15.5

A ball is travelling horizontally at 8 \displaystyle \text{m}{{\text{s}}^{-1}}, when it is hit. After being hit it initially travels upwards at 6 \displaystyle \text{m}{{\text{s}}^{-1}}. The mass of the ball is 300 grams. Find the magnitude and direction of the impulse on the ball.

Solution

The diagrams show the initial and final velocities of the ball and the unit vectors \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.

\displaystyle \begin{align} & \text{Initial Momentum }=\text{ }m\mathbf{u} \\ & =0\textrm{.}3\times 8\mathbf{i} \\ & =2\textrm{.}4\mathbf{i} \end{align}

\displaystyle \begin{align} & \text{Final Momentum }=m\mathbf{v} \\ & =0\textrm{.}3\times 6\mathbf{j} \\ & =1\textrm{.}8\mathbf{j} \end{align}

\displaystyle \begin{align} & \text{Impulse }=m\mathbf{v}-m\mathbf{u} \\ & =1\textrm{.}8\mathbf{j}-2\textrm{.}4\mathbf{i} \end{align}

The diagram shows the impulse.

The magnitude of the impulse, \displaystyle I, is found using Pythagoras:

\displaystyle I=\sqrt{{{1\textrm{.}8}^{2}}+{{2\textrm{.}4}^{2}}}=3\text{ Ns}

The angle, \displaystyle \alpha , can be found using trigonometry: