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Consider two cars, $a$ and $b$, which are currently at a resting state. The two cars are identical. Car $a$ starts up first, and accelerates at $2\, m/s^2$ until it reaches $100\, m/s$ and then promptly crashes into a wall. Car $b$ then does the same except it only accelerates until $50\, m/s$ and then promptly crashes into a wall. It is intuitive that car $a$ would cause more "damage" to the wall. However, using $F=ma$, the two car's forces are the same. Why is this?

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  • $\begingroup$ -1. Unclear. What forces? Please show your calculation that using $F=ma$ the two cars' forces are the same. $\endgroup$ – sammy gerbil Jul 13 '17 at 10:50
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The mistake is to consider that the "engine acceleration" of $2\, m/s^2$ is the cause of damage. The damage is actually caused by the deceleration during the collision. Assuming that the collision time is the same in both situations, the deceleration is higher for the car crashing at $100\, m/s$ and therefore causes more damage.

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  • $\begingroup$ @ Diracology Really what I'm getting at is 2 objects can be at different speeds and still exert the same force. Is this true? Also can you please clarify the definition of force. First time physics and sci-fi and fantasy movies do not seem to bode well... $\endgroup$ – Typical Highschooler Jul 12 '17 at 5:25
  • $\begingroup$ @TypicalHighschooler The force can indeed be calculated by $F=ma$. The point is that for car $a$ the average acceleration is $100/\Delta t$ whereas for car $b$ it is $50/\Delta t$. Hence acceleration and therefore the force are greater for car $a$. The acceleration of $2m/s^2$ has nothing to do with the collision. $\endgroup$ – Diracology Jul 12 '17 at 12:57
  • $\begingroup$ But the average acceleration is not 100/delta t and 50/delta t because the total distance is added on every time it accelerates. For example, say we have two identical cars in mass, a and b that both accelerate at 10 m/s. Car a drives 100 m forward in 4 seconds before crashing into a wall. 100m is obtained from 10 + 20 + 30 + 40 since it accelerates at 10m/s. Car b drives for 60 (changed for making example simpler) only for 3 seconds. 60 is obtained from 10 +20 + 30. The average acceleration for both cars from the start until the crash was 10m/s. Really confused here. @Diracology $\endgroup$ – Typical Highschooler Jul 12 '17 at 19:08
  • $\begingroup$ @TypicalHighschooler Average acceleration is defined as $\Delta v/\Delta t$ so what I said is correct. I think you still did not get the point of my answer. The acceleration to reach $100 m/s$ or $50m/s$ do not matter. What matters here is the acceleration to go from $100m/s$ and $50m/s$ to $0m/s$. That is the acceleration which relates to force causing the damage in the caras and that is the acceleration I am talking about. $\endgroup$ – Diracology Jul 12 '17 at 19:19
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The damage inflicted by a car crashing into a wall is caused by the net force applied by the wall on the car which is equal, according to the second law of Newton in terms of momentum to: $F_{net}=\left|\frac{\Delta p}{\Delta t}\right|$ where $\Delta p = m\Delta v$, given that the mass of the car doesn't change during the collision and $\Delta t$ is the collision time which depends on various factors such as the materials from which the car and the wall are made of and the arrangement of the parts which make the car (the degree of freedom which the car's components have). In the case of a car crashing into a wall the final velocity of the car is $0$ so $\Delta v=-v_{initial}$. This essentially means that the greater the initial collision velocity of the car with the wall is (given that the mass stays the same) and the shorter the time of the collision is, the greater the net force which is exerted on the car (and on the wall due to Newton's third law) will be, and the greater the damage inflicted on the car (and the wall) will be as well.

This has nothing to do with the car's acceleration but only with it's initial velocity at the time of collision. This is indeed intuitive as you said since a car crashing into a wall at a high constant speed (i.e. $0$ acceleration) would still take a serious amount of damage.

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