# Shortest collision time that will leave engine block intact?

## Homework Statement

In a crash test a solid concrete wall is
used to test crashworthiness. A particular
SUV weighs 338.183 kg and is accelerated to
33.3485 m/s before colliding head on with the
barrier.
What impulse must be applied is required
to stop the SUV?

This gives Impulse = 11277.896

The struggle comes with this part:

Through structural analysis of the SUV you
ﬁnd out that 21801 N will destroy the engine
block. If we ignore the friction between the
car and the ground and assume the force the
wall exerts a constant force on the car, What
is the shortest collision time that will leave
the engine block intact?

## Homework Equations

Impulse = net force * change in time

p(f) = p(i) + Impulse

## The Attempt at a Solution

So I assumed that a force of 21800.9N force would mean that the engine
is not broken. I do not know if this assumption holds. However, it seems to make
sense in my mind. So i figured if the wall exerts a constant force of 11277.896N
causing the car to have a momentum of 0; the change in time is:

21800.9N/11277.896N --> 1.933064465s

But this is apparently wrong.

gneill
Mentor

## Homework Statement

In a crash test a solid concrete wall is used to test crashworthiness. A particular SUV weighs 338.183 kg and is accelerated to 33.3485 m/s before colliding head on with the barrier.
What impulse must be applied is required to stop the SUV?

This gives Impulse = 11277.896 <---- units?

The struggle comes with this part:

Through structural analysis of the SUV you ﬁnd out that 21801 N will destroy the engine
block. If we ignore the friction between the car and the ground and assume the force the
wall exerts a constant force on the car, What is the shortest collision time that will leave
the engine block intact?

## Homework Equations

Impulse = net force * change in time

p(f) = p(i) + Impulse

## The Attempt at a Solution

So I assumed that a force of 21800.9N force would mean that the engine is not broken. I do not know if this assumption holds. However, it seems to make sense in my mind.
Yes, you can take that to be the maximum allowable force.
So i figured if the wall exerts a constant force of 11277.896N causing the car to have a momentum of 0; the change in time is:

21800.9N/11277.896N --> 1.933064465s

But this is apparently wrong.
Look at the units. N divided by N yields a result without units, not seconds.

Consider your first Relevant Equation, then look at my comment in red above.

I would also check 338.183 kg. It is much too little for an SUV. 3381.83 kg is more like it.

So I see that my initial move was naive..haha

I rethought the problem. Right before impact my momentum will be 11277.896N.

If I don't want to destroy my motor I need to only contact the wall (If i could be so lucky to choose) for a very small amount of time so that my Impulse = 21000.9 N.

So I get, p(i) = 11277.896 N and I have to stop moving so my final momentum p(f) = 0.

so I get:

0 = 11277.896 + (-21800.9N)t

So if this logic is correct i just rearrange and get t. I made the force negative because it's acting in the opposite direction of motion.

The kg is right...problem is clearly not realistic.

Does this seems right at all? The system we use docks points off of our total score if we try and fail..So in the spirit of not trying because of fear of failure a little assurance would be great. In other words, if you say No that is wrong then I can go back to the drawing board, else I already did the work so it's ok for me to know it's right.

gneill
Mentor
Make sure that you carry correct units through your calculations. What are the units of momentum and impulse?

Your last version is fine (post #4), except you forgot to include the units on the impulse value. What value do you calculate for the time?

Units of Impulse are kg*m/s --> N s; and momentum is N*s. Correct? and If the impulse value

11277.896N*s + (-21800.9N)t = 0 --> t = -11277.896N*s/-21800.9N = t (seconds)

(kg*m)/s * s*s/(kg*m) --> s

For some reason I believed that Force and Momentum had the same units...Now I see the error of my ways!

Thank you for the lending hand.

~Kevin

.517 s --> Correct. Thanks again

gneill
Mentor
You're welcome