# Velocity question

I’ve heard the stories of car crashes in which mom is riding in the passenger seat of a car and holding a small child in her lap. Upon impact the child becomes a projectile because mom isn’t physically capable of holding on to junior. My question is this: Is their a formula to determine how much said child weighs at the moment of impact based on the actual weight of the child and the speed at which the vehicle is travelling? I’m assuming it’s a mass/velocity calculation but I can’t seem to find it.
sly

It’s a mass/acceleration calculation: F = ma (force equals mass times acceleration)

A simple way of looking at this is that the horzontal force that mom will have to exert to hold onto junior is junior’s weight times the car’s deceleration in G.

What makes this less than simple is figuring out the peak deceleration. It does not depend only on the car’s initial velocity. It will have a lot to do with what the car hits and the distance over which its velocity changes.

Well, actually, the child’s weight never changes. Of course you knew that, I just can’t help being literal sometimes. What you really want to know is what the force of the child is relative to the mother trying to keep it from flying out the window. Let’s assume a 10lb baby, and a car travelling at 30mph (= 44feet/sec). Force equals mass times acceleration, or in this case, we’ll use decceleration. I think this works, but I’m kind of winging this as I go along and I’m sure someone else will be happy to correct me.
So, if the car’s goes from 30 to zero in say 20’, then v=v0 + at, or velocity (final) = velocity (initial) plus acceleration (deceleration) times time. Also (not doing this very streamlined, but work with me…), rate times time = distance. Our average velocity is (30+0)/2=15mph=22ft/sec. So t=d/r, or time=(20ft)/(22ft/sec)=.909 seconds to deccelerate from 30mph to zero. Back to the second equation! 0=44+a(.909), therefore a= -48.4 ft/sec^2.
Therefore, (if I got this right…) F = ma, so Force = 10lbs x (-48.4 ft/sec^2) = - 484 lbf (lbs force).
So, you’d have to be one strong Mama to hold your baby in even this mild of a crash. Ok, I’m sure it wouldn’t seem mild if you were in it, but I’m even suprised at the results.
There you go, now everyone can critique my logic and my flawed reasoning… Force = mass X acceleration

The force of child hitting the dashboard =
child’s weight (mass) X
(velocity of the car/duration of the collision)

Let’s say that the child weighs 10 kg.
Let’s discuss a head-on collision. I used to work for Ford in barrier test analysis and a typical barrier collision (which simulates a head-on collision into a car of equal weight) lasts about 0.3 seconds. Let’s also assume a speed of 50kph.

(10 kg)((50km/hr)/0.3 sec) =
(10 kg)((50km/hr)(1000m/km)(1 hr/3600 sec)/0.3 sec) =
463kg-m/sec^2

Now we have to convert kg-m/sec^2, which is units of force, to the equivalent mass under earth’s gravitational force.

(463kg-m/sec^2)/(9.8m/sec^2) =

47.2kg

This apparent mass is 4.7 times the child’s actual mass.

I use the term “apparent mass” to mean that the force with which the child hits the dashboard is equivalent to the force of gravity on a 47.2 kg mass.

I used metric because the math & physics are easier. In English units this means a 22 lb child would have a force equivalent to 104 lbs.

Assumptions.
x = 20 m distance through linear deceleration
m = 20 kg kid
Fg kid = 200 N
Vi = 13.5 m/s (~ 50kmph) initial velocity
Vf = 0 m/s final velocity

Compare the force involved to the force of gravity on the kid. Fg is 200N (20*10)

So in our case the car stops after 2.5 seconds and the mother experiences 91 N of force.

If we do a 100 kmph example t goes to 1.55 seconds and the mother experiences 338 N and I doubt throwing her arm across the seat is going help.

But hold on!

I’m not sure the analysis are fully correct. Remember, the mother is decelerating too! At the same rate. The above examples would give the force the baby would feel if it slammed into something; NOT the force the mother would have to exert to hold it.

The examples (except possibly CWG’s) are assuming that mother, baby and car maintain the same relationship with each other, and calculating the forces involved when that’s so. If the mother lets go and the baby slams into something, the calculation would be based on the rate of change in the baby’s velocity. In a real accident, this “seconday collision” is probably much more violent, which is the justification for seat/shoulder belts.

Note that the examples assume a constant deceleration. In the real world, this is unlikely. So the peak deceleration will be higher.

The child’s mass doesn’t change, if you really want to be literal. Weight is force, and does change.

Based on comments subsequent to my post, let me clarify my assumptions. I assumed that the mother is securely belted holding an unsecured child. In such a case, the force I calculated is the force with which the mother must hold onto the child to keep it from actually hitting the dashboard. This is roughly equivalent to a rowing weightlifting excercise. I can do about 10 reps on a T-bar loaded with 70 pounds. The average untrained female would probably not be able to hold onto a a child whose apparent forward force went from 0-100 pounds in 1/3 second (apparent in that the force is from the mother’s frame of reference). This is what the OP was asking for, IIRC

Well, if you really want to be literal the weight doesn’t change. Weight is force but it is specifically the force due to gravity, or local gravitational force times mass, so if gravity doesn’t change the weight doesn’t change. So your weight in a moving car and your weight in a crashing car and your weight in a stopping elevator are all the same. That’s why the term “weightless” is a misnomer for people in an orbiting spacecraft; they are still subject to gravity and definitely have weight. It’s just that both they and the spacecraft itself are both in free-fall. A cloud definitely has huge weight even thought it is floating in the air (see Cecil’s column on that).