# 0.2 Force, momentum and impulse  (Page 26/35)

 Page 26 / 35

## Newton's second law revisited

You have learned about Newton's Second Law of motion earlier in this chapter. Newton's Second Law describes the relationship between the motion of an object and the net force on the object. We said that the motion of an object, and therefore its momentum, can only change when a resultant force is acting on it. We can therefore say that because a net force causes an object to move, it also causes its momentum to change. We can now define Newton's Second Law of motion in terms of momentum.

Newton's Second Law of Motion (N2)

The net or resultant force acting on an object is equal to the rate of change of momentum.

Mathematically, Newton's Second Law can be stated as:

${F}_{net}=\frac{\Delta p}{\Delta t}$

## Impulse

Impulse is the product of the net force and the time interval for which the force acts. Impulse is defined as:

$\mathrm{Impulse}=F·\Delta t$

However, from Newton's Second Law, we know that

$\begin{array}{ccc}\hfill F& =& \frac{\Delta p}{\Delta t}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}F·\Delta t\phantom{\rule{4pt}{0ex}}& =& \Delta p\hfill \\ & =& \mathrm{Impulse}\hfill \end{array}$

Therefore,

$\mathrm{Impulse}=\Delta p$

Impulse is equal to the change in momentum of an object. From this equation we see, that for a given change in momentum, ${F}_{net}\Delta t$ is fixed. Thus, if ${F}_{net}$ is reduced, $\Delta t$ must be increased (i.e. a smaller resultant force must be applied for longer to bring about the same change in momentum). Alternatively if $\Delta t$ is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.

A 150 N resultant force acts on a 300 kg trailer. Calculate how long it takes this force to change the trailer's velocity from 2 m $·$ s ${}^{-1}$ to 6 m $·$ s ${}^{-1}$ in the same direction. Assume that the forces acts to the right.

1. The question explicitly gives

• the trailer's mass as 300 kg,
• the trailer's initial velocity as 2 m $·$ s ${}^{-1}$ to the right,
• the trailer's final velocity as 6 m $·$ s ${}^{-1}$ to the right, and
• the resultant force acting on the object

all in the correct units!

We are asked to calculate the time taken $\Delta t$ to accelerate the trailer from the 2 to 6 m $·$ s ${}^{-1}$ . From the Law of Momentum,

$\begin{array}{ccc}\hfill {F}_{net}\Delta t& =& \Delta p\hfill \\ & =& m{v}_{f}-m{v}_{i}\hfill \\ & =& m\left({v}_{f}-{v}_{i}\right).\hfill \end{array}$

Thus we have everything we need to find $\Delta t$ !

2. Choose right as the positive direction.

3. $\begin{array}{ccc}\hfill {F}_{net}\Delta t& =& m\left({v}_{f}-{v}_{i}\right)\hfill \\ \hfill \left(+150\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\right)\Delta \mathrm{t}& =& \left(300\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(\left(+6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)-\left(+2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)\right)\hfill \\ \hfill \left(+150\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\right)\Delta \mathrm{t}& =& \left(300\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(+4\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)\hfill \\ \hfill \Delta t& =& \frac{\left(300\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(+4\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)}{+150\phantom{\rule{0.166667em}{0ex}}\mathrm{N}}\hfill \\ \hfill \Delta t& =& 8\phantom{\rule{0.166667em}{0ex}}\mathrm{s}\hfill \end{array}$

It takes 8 s for the force to change the object's velocity from 2 m $·$ s ${}^{-1}$ to the right to 6 m $·$ s ${}^{-1}$ to the right.

A cricket ball weighing 156 g is moving at 54 km $·$ hr ${}^{-1}$ towards a batsman. It is hit by the batsman back towards the bowler at 36 km $·$ hr ${}^{-1}$ . Calculate

1. the ball's impulse, and
2. the average force exerted by the bat if the ball is in contact with the bat for 0,13 s.
1. The question explicitly gives

• the ball's mass,
• the ball's initial velocity,
• the ball's final velocity, and
• the time of contact between bat and ball

We are asked to calculate the impulse

$\mathrm{Impulse}=\Delta p={F}_{net}\Delta t$

Since we do not have the force exerted by the bat on the ball (F ${}_{net}$ ), we have to calculate the impulse from the change in momentum of the ball. Now, since

$\begin{array}{ccc}\hfill \Delta p& =& {p}_{f}-{p}_{i}\hfill \\ & =& m{v}_{f}-m{v}_{i},\hfill \end{array}$

we need the ball's mass, initial velocity and final velocity, which we are given.

2. Firstly let us change units for the mass

$\begin{array}{ccc}\hfill 1000\phantom{\rule{4pt}{0ex}}\mathrm{g}& =& 1\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\hfill \\ \hfill \mathrm{So},1\phantom{\rule{4pt}{0ex}}\mathrm{g}& =& \frac{1}{1000}\phantom{\rule{4pt}{0ex}}\mathrm{kg}\hfill \\ \hfill \therefore 156×1\phantom{\rule{4pt}{0ex}}\mathrm{g}& =& 156×\frac{1}{1000}\phantom{\rule{4pt}{0ex}}\mathrm{kg}\hfill \\ & =& 0,156\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\hfill \end{array}$

Next we change units for the velocity

$\begin{array}{ccc}\hfill 1\phantom{\rule{4pt}{0ex}}\mathrm{km}·{\mathrm{h}}^{-1}& =& \frac{1000\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{3\phantom{\rule{4pt}{0ex}}600\phantom{\rule{0.166667em}{0ex}}\mathrm{s}}\hfill \\ \hfill \therefore 54×1\phantom{\rule{4pt}{0ex}}\mathrm{km}·{\mathrm{h}}^{-1}& =& 54×\frac{1\phantom{\rule{0.166667em}{0ex}}000\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{3\phantom{\rule{0.166667em}{0ex}}600\phantom{\rule{0.166667em}{0ex}}\mathrm{s}}\hfill \\ & =& 15\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \end{array}$

Similarly, 36 km $·$ hr ${}^{-1}$ = 10 m $·$ s ${}^{-1}$ .

3. Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is ${v}_{i}$ = -15 m $·$ s ${}^{-1}$ , while the final velocity of the ball is ${v}_{f}$ = 10 m $·$ s ${}^{-1}$ .

4. Now we calculate the change in momentum,

$\begin{array}{ccc}\hfill p& =& {p}_{f}-{p}_{i}\hfill \\ & =& m{v}_{f}-m{v}_{i}\hfill \\ & =& m\left({v}_{f}-{v}_{i}\right)\hfill \\ & =& \left(0,156\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}\right)\left(\left(+10\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)-\left(-15\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\right)\right)\hfill \\ & =& +3,9\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}·\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ & =& 3,9\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}·\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\phantom{\rule{2pt}{0ex}}\mathrm{in the direction from batsman to bowler}\hfill \end{array}$
5. Finally since impulse is just the change in momentum of the ball,

$\begin{array}{ccc}\hfill \mathrm{Impulse}& =& \Delta p\hfill \\ & =& 3,9\phantom{\rule{0.166667em}{0ex}}\mathrm{kg}·\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\phantom{\rule{3pt}{0ex}}\mathrm{in the direction from the batsman to the bowler}\hfill \end{array}$
6. $\mathrm{Impulse}={F}_{net}\Delta t=\Delta p$

We are given $\Delta t$ and we have calculated the impulse of the ball.

$\begin{array}{ccc}\hfill {F}_{net}\Delta t& =& \mathrm{Impulse}\hfill \\ \hfill {F}_{net}\left(0,13\phantom{\rule{0.166667em}{0ex}}\mathrm{s}\right)& =& +3,9\phantom{\rule{0.166667em}{0ex}}\mathrm{N}·\phantom{\rule{0.166667em}{0ex}}\mathrm{s}\hfill \\ \hfill {F}_{net}& =& \frac{+3,9\phantom{\rule{0.166667em}{0ex}}\mathrm{N}·\phantom{\rule{0.166667em}{0ex}}\mathrm{s}}{0,13\phantom{\rule{0.166667em}{0ex}}\mathrm{s}}\hfill \\ & =& +30\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\hfill \\ & =& 30\phantom{\rule{0.166667em}{0ex}}\mathrm{N}\phantom{\rule{2.em}{0ex}}\mathrm{in the direction from batsman to bowler}\hfill \end{array}$

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