eli5: What is the difference in force when two individuals are going at an equal speed and collide vs one individual going at that same speed and colliding with a non moving individual?

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eli5: What is the difference in force when two individuals are going at an equal speed and collide vs one individual going at that same speed and colliding with a non moving individual?

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Force is mass times velocity/time.. if both objects are moving, then velocity is higher and time is lower making force much higher.

Depends a lot on how the person standing reacts to the impact.. Let’s break the problem down a bit. We’ll use cars since it’s easier to visualize due to the crumple zones collapsing.. All cars are identical and will be traveling at same speed.

1 car into immovable wall. – let this be our base line. The car will decelerate and crumple before rebounding a bit. Because the wall is immovable, the distance of the deceleration is the total deformation of the car. We’ll take this as our unit length of 1.

2 cars moving speed ‘v’ head on collision. – when the cars make contact, they start decelating at the same rate in opposite directions. This means the point of contact does not move during the collision for either car. Each car is experiencing the same distance for deceleration as above, so the impact will be the same forces as hitting the immovable wall.

1 car hitting a stopped car with no brakes applied. – This means the point of collision will move. The parked car will be accelerated backwards and collapse in the same direction as the moving car. The distance to decelerate the moving car is actually longer. This is important. Because the car has a further distance to come to a stop, it’s decelerating slower. The car that was stopped experiences equal and opposite forces.

Not sure ELI5 is the place for math, but I’ll give you the skinny: The amount of energy in the head on/wall collisions is 4 times that of one car hitting a stationary car!

This all works the same for people running into each other, but everything happens in much shorter distances since we don’t have crumple zones to collapse.

(this got long af sorry, there’s a short answer if you scroll to right near the end)

####Situation A:

Assume the two individuals are perfect mirrors of each other: they are exactly the same mass, m, and are travelling at exactly the same speed, (in opposite directions) towards each other, (+/-)v. From the conversation of momentum, we know they’ll both come to a complete rest upon collision. (Momentum = mass * velocity; the bodies have equal mass and equal but opposite velocities => they have equal but opposite momentums before collision => their momentum after collision is the sum of their initial momentums = 0; their final velocity is 0).

**Force:**

For each individual force is calculated using F = ma. Let the time taken for each individual to come to a stop after collision be t, then acceleration = (change in velocity) / time taken; a = (v-0)/t.

*So the force experienced by each individual = mv/t.*

**Energy:**

We calculate the energy released during the collision as the difference in the total kinetic energy before and after the collision, where KE = 0.5mv^2. After the collision: v=0, so KE=0. Before the collision: KE(i)= 2 * 0.5mv^2 = mv^2.

*So the energy released during the collision = mv^2*

####Situation B:

*Before collision:* individual A has mass m and velocity v, heading in the direction of the stationary individual B, who has mass m and velocity=0. Total momentum = mv. KE(i) = 0.5mv^2.

*After collision:* Assuming an inelastic collision, both individuals will move forward as one body (with mass 2m) after impact. From conservation of momentum, m(new) * v(new) = total momentum(before collision) => v(new) = mv / 2m = 0.5v.

**Force:**

For individual A, a = (v – 0.5v)/t = 0.5v/t. For individual B, a = (0.5v)/t = 0.5v/t.

*So the force experienced by each individual = 0.5mv/t.*

**Energy:**

After the collision: KE = 0.5(2m)(0.5v)^2 = 0.25mv^2. KE(released) = 0.5mv^2 – 0.25mv^2.

*So the energy released during the collision = 0.25mv^2*

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###Answer:
**Compared to a single moving individual colliding with a stationary individual, two individuals moving towards each other will each experience double the force on collision, and the collision will release 4 times as much energy.**
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(Possibly also a nice one to know: If an individual moving at the same speed collides with a wall (instead of another individual), the force experienced during that collision would be the same as in the collision with the second moving individual. This is a kind of fun demonstration of Newton’s third law (the “for every action there is an equal and opposite reaction” one). The wall’s mirroring of the single individual provides exactly the same response as in the collision between the two moving individuals. The collision with the wall would only release half as much energy as that between the two moving individuals though.)

Short version. If you have a car driving 30mph going into a wall the impact will have a certain amount of energy.

If you have 2 cars going head on 30mph then the collision would release exact same amount of the energy as a car hitting the wall at 60mph. That’s because 60mph is the relative speed between the cars.