When a net force acts on a body it will result in an acceleration which alters the
motion of the body. A large net force will cause a larger acceleration than a small net
force. The total change in motion of the object can be the same if the large and small
forces act for different time intervals. The combination of the force and time that it
acts is a useful quantity which leads us to define impulse.
Impulse is equal to the change in momentum of an object. From this equation we see,
that for a given change in momentum, \(\vec{F}_{net}\Delta t\) is fixed.
Thus, if \(\vec{F}_{net}\) 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.
The graphs below show how the force acting on a body changes with time.
The area under the graph, shaded in, represents the impulse of the body.
Worked example 15: Impulse and change in momentum
A \(\text{150}\) \(\text{N}\) resultant force
acts on a \(\text{300}\) \(\text{kg}\) trailer.
Calculate how long it takes this force to change the trailer's velocity
from \(\text{2}\) \(\text{m·s$^{1}$}\)
to \(\text{6}\) \(\text{m·s$^{1}$}\) in
the same direction. Assume that the forces acts to the right which is the direction
of motion of the trailer.
Identify what information is given and what is asked for
The question explicitly gives

the trailer's mass as \(\text{300}\) \(\text{kg}\),

the trailer's initial velocity as \(\text{2}\) \(\text{m·s$^{1}$}\) to the right,

the trailer's final velocity as \(\text{6}\) \(\text{m·s$^{1}$}\) to the right, and

the resultant force acting on the object.
We are asked to calculate the time taken \(\Delta t\) to accelerate the trailer from the \(\text{2}\) to \(\text{6}\) \(\text{m·s$^{1}$}\). From the Newton's second law,
\begin{align*}
\vec{F}_{net}\Delta t& = \Delta \vec{p} \\
& = m\vec{v}_{f}m\vec{v}_{i} \\
\Delta t& = \frac{m}{\vec{F}_{net}}\left(\vec{v}_{f}\vec{v}_{i}\right).
\end{align*}
Thus we have everything we need to find \(\Delta t\)!
Choose a frame of reference
Choose right as the positive direction.
Do the calculation and quote the final answer
\begin{align*}
\Delta t& = \frac{m}{\vec{F}_{net}}\left(\vec{v}_{f}\vec{v}_{i}\right) \\
\Delta \text{t}& = \left(\frac{300}{+150}\right)\left(\left(+6 \right)\left(+2 \right)\right) \\
\Delta \text{t}& = \left(\frac{300}{150}\right)\left(4 \right) \\
\Delta t& = \frac{\left(300\right)\left(+4 \right)}{150} \\
\Delta t& = \text{8}\text{ s}
\end{align*}
It takes \(\text{8}\) \(\text{s}\) for the force to change the object's velocity from \(\text{2}\) \(\text{m·s$^{1}$}\) to the right to \(\text{6}\) \(\text{m·s$^{1}$}\) to the right.
Worked example 16: Impulsive cricketers
A cricket ball weighing \(\text{156}\) \(\text{g}\)
is moving at \(\text{54}\) \(\text{km·hr$^{1}$}\)
towards a batsman. It is hit by the batsman back towards the bowler
at \(\text{36}\) \(\text{km·hr$^{1}$}\).
Calculate

the ball's impulse, and

the average force exerted by the bat if the ball is in contact with the bat for \(\text{0,13}\) \(\text{s}\).
Identify what information is given and what is asked for
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:
\[\text{Impulse}=\Delta \vec{p}=\vec{F}_{\text{net}}\Delta t\]
Since we do not have the force exerted by the bat on the ball ( \(\vec{F}_{\text{net}}\)), we have to calculate the impulse from the change in momentum of the ball. Now, since
\begin{align*}
\Delta \vec{p}& = \vec{p}_{f}\vec{p}_{i} \\
& = m\vec{v}_{f}m\vec{v}_{i},
\end{align*}
we need the ball's mass, initial velocity and final velocity, which we are given.
Convert to S.I. units
Firstly let us change units for the mass
\begin{align*}
\text{1 000} \text{g}& = 1 \text{kg} \\
\text{So,} 1 \text{g}& = \frac{1}{\text{1 000}} \text{kg} \\
\therefore 156\times 1 \text{g}& = 156\times \frac{1}{\text{1 000}} \text{kg} \\
& = \text{0,156} \text{kg}
\end{align*}
Next we change units for the velocity
\begin{align*}
1 \text{km}·{\text{h}}^{1}& = \frac{\text{1 000} \text{m}}{\text{3 600} \text{s}} \\
\therefore 54\times 1 \text{km}·{\text{h}}^{1}& = 54\times \frac{\text{1 000} \text{m}}{\text{3 600} \text{s}} \\
& = 15 \text{m·s$^{1}$}
\end{align*}
Similarly, \(\text{36}\) \(\text{km·hr$^{1}$}\) = \(\text{10}\) \(\text{m·s$^{1}$}\).
Choose a frame of reference
Let us choose the direction from the batsman to the bowler as the
positive direction. Then the initial velocity of the ball is
\(\vec{v}_{i}=\text{15}\text{ m·s$^{1}$}\), while the final
velocity of the ball is \(\vec{v}_{f}=+\text{10}\text{ m·s$^{1}$}\).
Calculate the momentum
Now we calculate the change in momentum,
\begin{align*}
\Delta \vec{p}& = \vec{p}_{f}\vec{p}_{i} \\
& = m\vec{v}_{f}m\vec{v}_{i} \\
& = m\left(\vec{v}_{f}\vec{v}_{i}\right) \\
& = \left(\text{0,156}\right)\left(\left(+10 \right)\left(15 \right)\right) \\
& = \text{+3,9} \\
& = \text{3,9}\text{ kg·m·s$^{1}$}~\text{in the direction from the batsman to the bowler}
\end{align*}
Determine the impulse
Finally since impulse is just the change in momentum of the ball,
\begin{align*}
\text{Impulse}& = \Delta \vec{p} \\
& =\text{3,9}\text{ kg·m·s$^{1}$}~\text{in the direction from the batsman to the bowler}
\end{align*}
Determine the average force exerted by the bat
\(\text{Impulse}=\vec{F}_{net}\Delta t=\Delta \vec{p}\)
We are given \(\Delta t\) and we have calculated the impulse of the ball.
\begin{align*}
\vec{F}_{net}\Delta t& = \text{Impulse} \\
\vec{F}_{net}\left(\text{0,13}\right)& = \text{+3,9} \\
\vec{F}_{net}& = \frac{\text{+3,9}}{\text{0,13}} \\
& = +30 \\
& = \text{30}\text{ N}~\text{in the direction from the batsman to the bowler}
\end{align*}
Worked example 17: Analysing a force graph
Analyse the Force vs. time graph provided and answer the following questions:
 What is the impulse for the interval
\(\text{0}\) \(\text{s}\) to
\(\text{3}\) \(\text{s}\)?
 What is the impulse for the interval
\(\text{3}\) \(\text{s}\) to
\(\text{6}\) \(\text{s}\)?
 What is the change in momentum for the interval
\(\text{0}\) \(\text{s}\) to
\(\text{6}\) \(\text{s}\)?
 What is the impulse for the interval
\(\text{6}\) \(\text{s}\) to
\(\text{20}\) \(\text{s}\)?
 What is the impulse for the interval
\(\text{0}\) \(\text{s}\) to
\(\text{20}\) \(\text{s}\)?
Identify what information is given and what is being asked for
A graph of force versus time is provided. We are asked to determine both impulse
and change in momentum from it.
We know that the area under the graph is the impulse and we can relate impulse to
change in momentum with the impulsemomentum theorem.
We need to calculate the area under the graph for the various intervals to determine impulse
and then work from there.
Impulse for interval \(\text{0}\) \(\text{s}\) to
\(\text{3}\) \(\text{s}\)
We need to calculate the area of the shaded portion under the graph. This is a triangle with
a base of \(\text{3}\) \(\text{s}\) and a height of
\(\text{3}\) \(\text{N}\) therefore:
\begin{align*}
\text{Impulse} &= \frac{1}{2}bh \\
& = \frac{1}{2}(3)(3) \\
& = \text{4,5}\text{ N·s}
\end{align*}
The impulse is \(\text{4,5}\) \(\text{N·s}\) in the positive direction.
Impulse for interval \(\text{3}\) \(\text{s}\) to
\(\text{6}\) \(\text{s}\)
We need to calculate the area of the shaded portion under the graph. This is a triangle with
a base of \(\text{3}\) \(\text{s}\) and a height of
\(\text{3}\) \(\text{N}\). Note that the force has a
negative value so is pointing in the negative direction.
\begin{align*}
\text{Impulse} &= \frac{1}{2}bh \\
& = \frac{1}{2}(3)(3) \\
& = \text{4,5}\text{ N·s}
\end{align*}
The impulse is \(\text{4,5}\) \(\text{N·s}\) in the negative direction.
What is the change in momentum for the interval \(\text{0}\) \(\text{s}\) to
\(\text{6}\) \(\text{s}\)
From the impulsemomentum theorem we know that that impulse is equal to the change in
momentum. We have worked out the impulse for the two subintervals making up
\(\text{0}\) \(\text{s}\) to
\(\text{6}\) \(\text{s}\). We can sum them to find the
impulse for the total interval:
\begin{align*}
\text{impulse}_{06} & = \text{impulse}_{03} + \text{impulse}_{36} \\
&= (\text{4,5})+(\text{4,5}) \\
&= \text{0}\text{ N·s}
\end{align*}
The positive impulse in the first 3 seconds is exactly opposite to the impulse in the second
3 second interval making the total impulse for the first 6 seconds zero:
\[\text{impulse}_{06} = \text{0}\text{ N·s}\]
From the impulsemomentum theorem we know that:
\[\Delta \vec{p} = \text{impulse} = \text{0}\text{ N·s}\]
What is the impulse for the interval \(\text{6}\) \(\text{s}\) to
\(\text{20}\) \(\text{s}\)
We need to calculate the area of the shaded portion under the graph.
This is divided into two areas, \(\text{6}\) \(\text{s}\) to \(\text{12}\) \(\text{s}\) and \(\text{12}\) \(\text{s}\) to \(\text{20}\) \(\text{s}\), which we need to sum to get the
total impulse.
\begin{align*}
\text{Impulse}_{612} &= (6)(3) \\
& = \text{18}\text{ N·s}
\end{align*}\begin{align*}
\text{Impulse}_{1220} &= (8)(2) \\
& = +\text{16}\text{ N·s}
\end{align*}
The total impulse is the sum of the two:
\begin{align*}
\text{Impulse}_{620} &= \text{Impulse}_{612} + \text{Impulse}_{1220} \\
&= (18) + (16) \\
&= \text{2}\text{ N·s}
\end{align*}
The impulse is \(\text{2}\) \(\text{N·s}\) in the negative direction.
What is the impulse of the entire period
The impulse is \(\text{2}\) \(\text{N·s}\) in the negative direction.
Worked example 18: Car chase [Excerpt from NSC 2011 Paper 1]
A patrol car is moving on a straight horizontal road at a velocity of \(\text{10}\) \(\text{m·s$^{1}$}\) east. At the same time a thief in a car ahead of him is driving at a velocity of \(\text{40}\) \(\text{m·s$^{1}$}\) in the same direction.
\(v_{PG}\): velocity of the patrol car relative to the ground
\(v_{TG}\): velocity of the thief's car relative to the ground
Questions 1 and 2 from the original version in 2011 Paper 1 are no longer
part of the curriculum.

While travelling at \(\text{40}\) \(\text{m·s$^{1}$}\), the thief's car of mass \(\text{1 000}\) \(\text{kg}\), collides headon with a truck of mass \(\text{5 000}\) \(\text{kg}\) moving at \(\text{20}\) \(\text{m·s$^{1}$}\). After the collision, the car and the truck move together. Ignore the effects of friction.
State the law of conservation of linear momentum in words.
(2 marks)

Calculate the velocity of the thief's car immediately after the collision.
(6 marks)

Research has shown that forces greater than 85 000 N during collisions may cause fatal injuries. The collision described above lasts for \(\text{0,5}\) \(\text{s}\).
Determine, by means of a calculation, whether the collision above could result in a fatal injury.
(5 marks)
[TOTAL: 13 marks]
Question 1
The total (linear) momentum remains constant (OR is conserved OR does not change) in an isolated (OR in a closed system OR in the absence of external forces).
(2 marks)
Question 2
Option 1:
Taking to the right as positive
\begin{align*}
\sum p_{\text{before}} & = \sum p_{\text{after}} \\
(\text{1 000})(\text{40})+(\text{5 000})(\text{20}) & = (\text{1 000}+\text{5 000})v_f \\
v_f & = \text{10 m}\cdot\text{s}^{1}\\
& = \text{10 m}\cdot\text{s}^{1} \quad \text{left OR west}
\end{align*}
Option 2:
Taking to the right as positive
\begin{align*}
\Delta p_{\text{car}}& =  \Delta p_{\text{truck}} \\
m_{\text{car}}(v_f v_{i,\text{car}}) & =  m_{\text{truck}}(v_f v_{i,\text{truck}}) \\
(\text{1 000})(v_f(\text{40})) & = (\text{5 000})(v_f (\text{20})) \\
\text{6 000}v_f & = \text{60 000} \\
\therefore v_f & = \text{10 m}\cdot\text{s}^{1} \\
\therefore v_f & = \text{10 m}\cdot\text{s}^{1} \quad \text{left OR west}
\end{align*}
(6 marks)
Question 3
Option 1:
Force on the car: (Taking to the right as positive)
\begin{align*}
F_{\text{net}} \Delta t & = \Delta p = mv_f  mv_i \\
F_{\text{net}} (\text{0,5}) & = (\text{1 000})(\text{10}\text{40}) \\
\therefore F_{\text{net}} & = 10^5\text{ N} \\
& \text{OR} \\
\therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
OR
Force on the car: (Taking to the left as positive)
\begin{align*}
F_{\text{net}} \Delta t & = \Delta p = mv_f  mv_i \\
F_{\text{net}} (\text{0,5}) & = (\text{1 000})(\text{10}(40)) \\
\therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
Option 2:
Force on the truck: (Taking to the right as positive)
\begin{align*}
F_{\text{net}} \Delta t & = \Delta p = mv_f  mv_i \\
F_{\text{net}} (\text{0,5}) & = (\text{5 000})(10(20)) \\
\therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
OR
Force on the truck: (Taking to the left as positive)
\begin{align*}
F_{\text{net}} \Delta t & = \Delta p = mv_f  mv_i \\
F_{\text{net}} (\text{0,5}) & = (\text{1 000})(\text{10}\text{20}) \\
\therefore F_{\text{net}} & = 10^5\text{ N} \\
& \text{OR} \\
\therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
Option 3:
Force on the car: (Taking to the right as positive)
\begin{align*}
v_f & = v_i + a \Delta t \\
10 & = \text{40} + a (\text{0,5})\\
\therefore a & = \text{100 m}\cdot\text{s}^{2}\\
F_{\text{net}} & = ma \\
& = (\text{1 000})(100) \\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
OR
Force on the car: (Taking to the left as positive)
\begin{align*}
v_f & = v_i + a \Delta t \\
\text{10} & = 40 + a (\text{0,5})\\
\therefore a & = \text{100 m}\cdot\text{s}^{2}\\
F_{\text{net}} & = ma \\
& = (\text{1 000})(\text{100}) \\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
Option 4:
Force on the truck: (Taking to the right as positive)
\begin{align*}
v_f & = v_i + a \Delta t \\
10 & = 20 + a (\text{0,5})\\
\therefore a & = \text{20 m}\cdot\text{s}^{2}\\
F_{\text{net}} & = ma \\
& = (\text{5 000})(\text{20}) \\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
OR
Force on the truck: (Taking to the left as positive)
\begin{align*}
v_f & = v_i + a \Delta t \\
\text{10} & = \text{20} + a (\text{0,5})\\
\therefore a & = \text{20 m}\cdot\text{s}^{2}\\
F_{\text{net}} & = ma \\
& = (\text{5 000})(20) \\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\
\therefore F_{\text{net}} & > \text{85 000 N}
\end{align*}
Yes, the collision is fatal.
(5 marks)
[TOTAL: 13 marks]