B Acceleration regards F=ma

[paulb203]

TL;DR Summary

When calculating the acceleration of a car driving into a wall in order to calculate the force via F=ma, does it matter whether the car was accelerating or not prior to impact?

I'm thinking it's irrelevant, whether the car was driving at a constant velocity, or accelerating, prior to impact, as I'm thinking the v and u that matter for the F=ma calculation are; the v of 0m/s once the car has come to a stop, and the v of that on impact. But I'm not sure.

 

[PeroK]

TL;DR Summary: When calculating the acceleration of a car driving into a wall in order to calculate the force via F=ma, does it matter whether the car was accelerating or not prior to impact?

I'm thinking it's irrelevant, whether the car was driving at a constant velocity, or accelerating, prior to impact, as I'm thinking the v and u that matter for the F=ma calculation are; the v of 0m/s once the car has come to a stop, and the v of that on impact. But I'm not sure.

If the car is accelerating towards the wall, then there must be a net force on the car towards the wall. If that forces vanishes the instant the car hits the wall, then it does not affect the collision. If, however, it continues during the collision, then it does affect the contact force. An example is a vertical pile-driver, where the weight of the pile-driver is significant in terms of the force it exerts. I.e. the force of gravity affcets the collision in addition to the speed of the pile-driver on impact.

 

[russ_watters]

I agree with the above, however it is also likely that the car's acceleration towards the wall is a small fraction of the deceleration of the impact. So for most purposes it can probably be ignored.

 

[renormalize]

...celebration of the impact.

Like watching the destruction derby event at the county fair: the crowd goes wild!

 

[russ_watters]

Like watching the destruction derby event at the county fair: the crowd goes wild!

Autocorrect fail fixed.

 

[paulb203]

Thanks, guys.

I’ll try an example.

Car A

Mass; 1000kg

Constant v; 15m/s

Time(from hitting the wall to stopping); 0.1s

Q. To establish the acceleration for the F=ma calculation do I do;

A=(v-u)/t

A=(0m/s-15m/s)/t

A= -15/0.1

A= -150m/s/s ?

**

Car B

Mass; 1000kg

A= 5m/s/s

V on impact; 15m/s

Time (from hitting the wall to stopping); 0.1s

Q. To establish the acceleration for the F=ma calculation for Car B do I do the same as I did for Car A;

A=(v-u)/t

A=(0m/s-15m/s)/t

A= -15/0.1

A= -150m/s/s ?

 

[jbriggs444]

Car A

Mass; 1000kg

Constant v; 15m/s

Time(from hitting the wall to stopping); 0.1s

Q. To establish the acceleration for the F=ma calculation do I do;

A=(v-u)/t

A=(0m/s-15m/s)/t

A= -15/0.1

A= -150m/s/s ?

Yes. To come to a stop in 0.1 seconds from an initial velocity of 15 m/s requires an average acceleration of -150 m/s/s.

We can compute the required average collision force as $F_\text{collision} = mA_\text{collision} = 5000 \times -150 = -750000$ N

Car B

Mass; 1000kg

A= 5m/s/s

V on impact; 15m/s

Time (from hitting the wall to stopping); 0.1s

Q. To establish the acceleration for the F=ma calculation for Car B do I do the same as I did for Car A;

A=(v-u)/t

You are going to confuse yourself if you use the same variable name "A" to denote both the acceleration rate during the collision and the acceleration rate prior to the collision.

Let us use $A_\text{prior}$ to denote the prior acceleration rate.

This means that someone is applying a forward force of $F_\text{prior} = mA_\text{prior} = 5000 \times 5 = 25000$ N prior to and during the collision.

A=(0m/s-15m/s)/t

A= -15/0.1

A= -150m/s/s ?

The required acceleration is the same. But what average collision force is now required to achieve that acceleration?

The net force is still going to be -750000 N, But with the engine still running and the tires still spinning and providing 25000 N of forward force, the collision will need to produce -775000 N of retarding force to result in the same net force and, consequently, the same deceleration profile.

This is somewhat unrealistic. How would the wall know that it needs to produce more force against car B than against car A?

The collision time is in the right ball park. Assuming constant deceleration, we have an average speed of 7.5 m/s during a collision of 0.1 s duration. So about 75 centimeters of crush.

 

[paulb203]

Yes. To come to a stop in 0.1 seconds from an initial velocity of 15 m/s requires an average acceleration of -150 m/s/s.

We can compute the required average collision force as $F_\text{collision} = mA_\text{collision} = 5000 \times -150 = -750000$ N

You are going to confuse yourself if you use the same variable name "A" to denote both the acceleration rate during the collision and the acceleration rate prior to the collision.

Let us use $A_\text{prior}$ to denote the prior acceleration rate.

This means that someone is applying a forward force of $F_\text{prior} = mA_\text{prior} = 5000 \times 5 = 25000$ N prior to and during the collision.


The required acceleration is the same. But what average collision force is now required to achieve that acceleration?

The net force is still going to be -750000 N, But with the engine still running and the tires still spinning and providing 25000 N of forward force, the collision will need to produce -775000 N of retarding force to result in the same net force and, consequently, the same deceleration profile.

This is somewhat unrealistic. How would the wall know that it needs to produce more force against car B than against car A?

The collision time is in the right ball park. Assuming constant deceleration, we have an average speed of 7.5 m/s during a collision of 0.1 s duration. So about 75 centimeters of crush.

Thanks.
For Car A I had -150,000N of F, but I think you've put 5000kg in for the mass instead of 1000kg.

 

[jbriggs444]

Thanks.
For Car A I had -150,000N of F, but I think you've put 5000kg in for the mass instead of 1000kg.

Yes, sorry.