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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(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.)
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