9.2 Impulse and collisions  (Page 4/10)

Solution

1. Define the + x -direction to be the direction the car is initially moving. We know
   
   $\overrightarrow{J}=\overrightarrow{F}\text{Δ}t$
   
   and
   
   $\overrightarrow{J}=m\text{Δ}\overrightarrow{v}.$
   
   Since J is equal to both those things, they must be equal to each other:
   
   $\overrightarrow{F}\text{Δ}t=m\text{Δ}\overrightarrow{v}.$
   
   We need to convert this weight to the equivalent mass, expressed in SI units:
   
   $\frac{860\text{N}}{9.8{\text{m/s}}^2}=87.8\text{kg}.$
   
   Remembering that $\text{Δ}\overrightarrow{v}={\overrightarrow{v}}_{\text{f}}-{\overrightarrow{v}}_{\text{i}}$ , and noting that the final velocity is zero, we solve for the force:
   
   $\overrightarrow{F}=m\frac{0-v_{\text{i}}\widehat{i}}{\text{Δ}t}=\left(87.8\text{kg}\right)\left(\frac{\text{−}\left(27\text{m}\text{/}\text{s}\right)\widehat{i}}{2.5\text{s}}\right)=\text{−}\left(948\text{N}\right)\widehat{i}.$
   
   The negative sign implies that the force slows him down. For perspective, this is about 1.1 times his own weight.
2. Same calculation, just the different time interval:
   
   $\overrightarrow{F}=\left(87.8\text{kg}\right)\left(\frac{\text{−}\left(27\text{m}\text{/}\text{s}\right)\widehat{i}}{0.20\text{s}}\right)=\text{−}\left(\mathrm{11,853}\text{N}\right)\widehat{i}$
   
   which is about 14 times his own weight. Big difference!

Significance

Effect of impulse

Since an impulse is a force acting for some amount of time, it causes an object’s motion to change. Recall [link] :

Because $m\overrightarrow{v}$ is the momentum of a system, $m\text{Δ}\overrightarrow{v}$ is the change of momentum $\text{Δ}\overrightarrow{p}$ . This gives us the following relation, called the impulse-momentum theorem    (or relation).

The impulse-momentum theorem is depicted graphically in [link] .

There are two crucial concepts in the impulse-momentum theorem:

1. Impulse is a vector quantity; an impulse of, say, $\text{−}\left(10\text{N}·\text{s}\right)\widehat{i}$ is very different from an impulse of $+\left(10\text{N}·\text{s}\right)\widehat{i}$ ; they cause completely opposite changes of momentum.
2. An impulse does not cause momentum; rather, it causes a change in the momentum of an object. Thus, you must subtract the final momentum from the initial momentum, and—since momentum is also a vector quantity—you must take careful account of the signs of the momentum vectors.

The most common questions asked in relation to impulse are to calculate the applied force, or the change of velocity that occurs as a result of applying an impulse. The general approach is the same.

Moving the Enterprise

“Mister Sulu, take us out; ahead one-quarter impulse.” With this command, Captain Kirk of the starship Enterprise ( [link] ) has his ship start from rest to a final speed of $v_{\text{f}}=1\text{/}4\left(3.0\times {10}^8\text{m/s}\right)$ . Assuming this maneuver is completed in 60 s, what average force did the impulse engines apply to the ship?

Strategy

We are asked for a force; we know the initial and final speeds (and hence the change in speed), and we know the time interval over which this all happened. In particular, we know the amount of time that the force acted. This suggests using the impulse-momentum relation. To use that, though, we need the mass of the Enterprise . An internet search gives a best estimate of the mass of the Enterprise (in the 2009 movie) as $2\times {10}^9\text{kg}$ .

Solution

Because this problem involves only one direction (i.e., the direction of the force applied by the engines), we only need the scalar form of the impulse-momentum theorem [link] , which is

$\text{Δ}p=J$

with

$\text{Δ}p=m\text{Δ}v$

and

$J=F\text{Δ}t.$

Equating these expressions gives

$F\text{Δ}t=m\text{Δ}v.$

Solving for the magnitude of the force and inserting the given values leads to

$F=\frac{m\text{Δ}v}{\text{Δ}t}=\frac{\left(2\times {10}^9\text{kg}\right)\left(7.5\times {10}^7\text{m/s}\right)}{60\text{s}}=2.5\times {10}^{15}\text{N}.$

Significance

This is an unimaginably huge force. It goes almost without saying that such a force would kill everyone on board instantly, as well as destroying every piece of equipment. Fortunately, the Enterprise has “inertial dampeners.” It is left as an exercise for the reader’s imagination to determine how these work.

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