FHSST Physics/Momentum/Impulse

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Momentum
Definition - Momentum of a System - Change in Momentum - Properties - Impulse - Important Quantities, Equations, and Concepts


At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection.

Consider an object of mass m moving with constant acceleration \overrightarrow{a}. During a time \Deltat the object's velocity changes from an initial velocity \overrightarrow{u} to a final velocity \overrightarrow{v} (refer to Figure 6.3). We know from Newton's First Law that there must be a resultant force \overrightarrow{F}_{Res} acting on the object.

Fhsst mom4.png
Figure 6.3: An object under the action of a resultant force.

Starting from Newton's Second Law,

\begin{matrix}\overrightarrow{F}_{Res}&=&m\overrightarrow{a}\\&=&m(\frac{\overrightarrow{v}-\overrightarrow{u}}{\Delta t})\qquad\qquad\rm {since}\qquad\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{\Delta t}\\&=&\frac{m\overrightarrow{v}-m\overrightarrow{u}}{\Delta t}\\&=&\frac{\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}}{\Delta t}\\&=&\frac{\Delta \overrightarrow{p}}{\Delta t}\end{matrix}

This alternative form of Newton's Second Law is called the Law of Momentum.

Mathematically,

\overrightarrow{F}_{Res} =\frac{\Delta \overrightarrow{p}}{\Delta t}
\overrightarrow{F}_{Res}  : resultant force (N + direction)
\Delta \overrightarrow{p}  : change in momentum (kg.m.s^{-1} + direction)
\Deltat  : time over which \overrightarrow{F}_{Res} acts (s)

Rearranging the Law of Momentum,

\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \Delta \overrightarrow{p}.\end{matrix}

The product \overrightarrow{F}_{Res}\Delta t is called impulse,

\mathrm{Impulse} \equiv \overrightarrow{F}_{Res}\Delta t = \Delta\overrightarrow{p}

From this equation we see, that for a given change in momentum,\overrightarrow{F}_{Res}\Delta t is fixed. Thus, if FRes is reduced, \Delta t must be increased (i.e. the resultant force must be applied for longer). 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.

Worked Example 36 Impulse and Change in momentum[edit]

Question: A 150   N resultant force acts on a 300   kg object. Calculate how long it takes this force to change the object's velocity from 2\ m.s^{-1}\ to\ the\ right to 6\ m.s^{-1}\ to\ the\ right

Answer:

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

  • the object's mass,
  • the object's initial velocity,
  • the object's final velocity, and
  • the resultant force acting on the object

all in the correct units!

Step 2 :

What is being asked? We are asked to calculate the time taken \Delta t to accelerate the object from the given initial velocity to final velocity. From the Law of Momentum,

\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \Delta \overrightarrow{p} \\&=& m\overrightarrow{v}-m\overrightarrow{u}\\&=& m(\overrightarrow{v}-\overrightarrow{u}).\end{matrix}

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

Step 3 :

First we choose a positive direction. Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object's velocity increases in this direction. Let us then choose right as the positive direction.

Step 4 :

Substituting,

Right is the positive direction

\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& m(\overrightarrow{v}-\overrightarrow{u})\\(+150N)\Delta t &=& (300kg)((+6\frac{m}{s})-(+2\frac{m}{s}))\\(+150N)\Delta t &=& (300kg)(+4\frac{m}{s})\\\Delta t &=& \frac{(300kg)(+4\frac{m}{s})}{+150N}\\\Delta t &=& 8s\end{matrix}

Worked Example 37 Calculating Impulse[edit]

Question: A cricket ball weighing 156 g is moving at 54 km/h towards a batsman. It is hit by the batsman back towards the bowler at 36\ km/h. Calculate i) the ball's impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0.13 s.

Answer:

Step 1 :

Analyse the question to determine what is given. 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

all except the time in the wrong units!

Answer to (i):

Step 2 :

What is being asked? We are asked to calculate the impulse

\begin{matrix}\mathrm{Impulse} = \Delta\overrightarrow{p} =\overrightarrow{F}_{Res}\Delta t. \end{matrix}

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

\begin{matrix}\Delta\overrightarrow{p} &=&\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}\\&=& m\overrightarrow{v} - m\overrightarrow{u},\end{matrix}

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

Step 3 : Firstly let us change units for the mass

\begin{matrix}1000g &=& 1kg\\1 &=& \frac{1kg}{1000g}\\156g \times 1 &=& 156g \times \frac{1kg}{1000g}\\&=& 0.156kg\end{matrix}

Step 4 :

Next we change units for the velocity

\begin{matrix}1km &=& 1000 m\\1 &=& \frac{1000m}{1km}\end{matrix}
\begin{matrix}3600s &=& 1hr\\1 &=& \frac{1hr}{3600 s}\end{matrix}
\begin{matrix}54\frac{km}{hr}\times 1\times1 &=&54\frac{km}{hr}\times\frac{1000m}{1km}\times\frac{1hr}{3600 s}\\&=& 15\frac{m}{s}\end{matrix}
\begin{matrix}36\frac{km}{hr}\times 1\times1 &=&36\frac{km}{hr}\times\frac{1000m}{1km}\times\frac{1hr}{3600 s}\\&=& 10\frac{m}{s}\end{matrix}

Step 5 :

Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is \overrightarrow{u}=-15\ m.s^{-1}, while the final velocity of the ball is \overrightarrow{v}=+10\ m.s^{-1}

Step 6 :

Now we calculate the change in momentum,

Direction from batsman to bowler is the positive direction

\begin{matrix}\Delta\overrightarrow{p} &=&\overrightarrow{p}_{final}-\overrightarrow{p}_{initial}\\&=& m\overrightarrow{v}-m\overrightarrow{u}\\&=& m(\overrightarrow{v}-\overrightarrow{u})\\&=&(0.156kg)((+10\ m.s^{-1})-(-15\ m.s^{-1}))\\&=& +3.9\ kg.m.s^{-1}\\&=& 3.9\ kg.m.s^{-1}\ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}

where we remembered in the last step to include the direction of the change in momentum in words.

Step 7 :

Finally since impulse is just the change in momentum of the ball,

\begin{matrix}\mathrm{Impulse} &=& \Delta\overrightarrow{p}\\&=& 3.9\ \mbox{kg} \cdot \mbox{m} \cdot \mbox{s}^{-1}\\&=& 3.9\ \mbox{N} \cdot \mbox{s} \ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}

Answer to (ii):

Step 8 :

What is being asked? We are asked to calculate the average force exerted by the bat on the ball, \overrightarrow{F}_{Res}. Now,

\begin{matrix}\mathrm{Impulse} = \overrightarrow{F}_{Res}\Delta t = \Delta\overrightarrow{p}.\end{matrix}

We are given \Delta t and we have calculated the change in momentum or impulse of the ball in part (i)!

Step 9 :

Next we choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then substituting,

Direction from batsman to bowler is the positive direction

\begin{matrix}\overrightarrow{F}_{Res}\Delta t &=& \mathrm{Impulse}\\\overrightarrow{F}_{Res}(0.13s) &=& +3.9\frac{kg.m}{s}\\\overrightarrow{F}_{Res} &=& \frac{+3.9\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}}}{0.13 \mbox{ s}}\\&=& 30 \mbox{ N} \ \textbf{ in\ the\ direction\ from\ batsman\ to\ bowler}\end{matrix}

where we remembered in the final step to include the direction of the force in words.