# What causes more damage, a head-on car crash at 30km/h or one car at 60km/h crashing into a stationary one?

Which will cause a greater impact?

• Two identical cars colliding head on (in opposite directions), each going at $30\;\mathrm{km/h}$

OR

• One car travelling at $60\;\mathrm{km/h}$ colliding into a stationary car (for the same two cars).

Also, why is this so?

• Sorry, I forgot to mention. The two cars are identical. – John Smith Jul 26 '16 at 13:21
• What exactly is your definition of "greater impact"? – ACuriousMind Jul 27 '16 at 13:07

Car collision "damage" usually goes with the energy in the zero momentum frame.

In both cases that is (since in the zero momentum frame, the two cases are equivalent, assuming the masses of the cars are equal): $$E_1 = 2 \times \frac{1}{2} m v_{rel}^2 = m \left( 30 \frac{km}{h} \right)^2$$ Therefore a priori there is no difference between the two situations.

One may have to consider other factors. E.g. in the heads on collision, the cars would be roughly stationary afterwards (assuming a fully inelastic collision, which is reasonable since cars are quite "squashable" due to their safety features). In the other case there would still be motion which could result in a secondary crash.

in this answer im not substituting values with proper units as it is all just about comparrison

there are two ways of refering impact: force and impulse and for each body impulse is different

impulse on an body = change in its momentum

ultimately if you measure impact as impulse,

• first case both car have same change in velocity (30 to 0) if masses are equal impulse will be equal
• in second case car have change in velocity (60 to 0) while wall have (0 to 0) assuming the wall is being stationary after hit. so thee car have double the impulse of what the car(assuming same masses) had in above situation, that is the impulse is shared between cars in above situation

if you measure impact as force imparted on the body, a time variable comes, the impact time and that force is equal to M(V-U)/t where t is time of impact:

• in first case the force exerted by car on to each other is M(30-0)/t=30M/t. imagine you are in a car after impact your car gives force 30M/t to the other car, due to newton third law you get the same amount of force as reaction force in addition to the force given by opposite car's deceleration so in total you get 30M/t + 30M/t = 60M/t

• in the second u r dashing onto the wall exerting a force of M(60-0)/t=60M/t due to newton third law you get the same amount of force as reaction force on u, here unlike the above situation there is no additional force from wall as it did not accelerate or decelerate. so the force on each other is 60M/t

conclusion: answer for your question depends on how you define the impact, if you are talking about impulse in the second case the driver suffer 2x the imulse suffered by drivers in 1st case. if u are asking about force in both situations forces imparted are same

@Numrok 's analysis makes certain assumptions which are then mentioned in the last paragraph of the answer.

For simplicity assume that the speed of the cars is measured in m/s rather than km/hr.

In the head on collision with both cars of the same mass and speed the final kinetic energy will be zero if the cars interlock as a result of the collision.
So the loss of kinetic energy $(= 2 \times \frac 1 2 \; m\; 30^2)$, which is a measure of the damage done due to permanent deformation, is twice the initial kinetic energy of one of the cars travelling at 30 m/s $(=\frac 1 2 \; m\; 30^2)$.
If the cars do not interlock then the loss in kinetic energy will be less.

In the other case assuming that the handbrake is off in the stationary car and the cars interlock after the collision the cars move off together at 30 /s with a total kinetic energy of $2 \times \frac 1 2 \; m\; 30^2$ but the initial kinetic energy was $\frac 1 2 \; m\; 60^2 = 4 \times \frac 1 2 \; m\; 30^2$ so the loss of kinetic energy is $2 \times \frac 1 2 \; m\; 30^2$ which is the same as before.

However with the brake on and/or friction present the loss of kinetic energy with both cars finishing up stationary is $\frac 1 2 \; m\; 60^2 = 4 \times \frac 1 2 \; m\; 30^2$ which is more than the head on collision with both cars initially moving at $30$ m/s.

Given that the $2 \times 30$ m/s collision will probably lose less than the maximum $2 \times \frac 1 2 \; m\; 30^2$ amount of kinetic energy whereas the $0+ 60$ m/s collision will lose less than $4 \times \frac 1 2 \; m\; 30^2$ but more than $2 \times \frac 1 2 \; m\; 30^2$ amount of kinetic energy I would say that the collision with the $60$ m/s car hitting the stationary car will probably do more damage to the cars.

As pointed out by @RahulJA the "damage" done to the occupants of the cars is much more difficult to quantify as their acceleration is a factor which must be considered.

I would think the second situation causes more damage.

The total energy of the first situation is $$2\cdot\frac 12 m v^2$$ The second situation total energy is $$\frac 12 m (2v)^2=4\cdot\frac 12 m v^2$$ After the collision, car plastic deformation energy can be approximated by the total kinetic energy. Well you can argue there are other way to consume the kinetic energy. Those are second order in magnitude.

## protected by Qmechanic♦Jul 27 '16 at 11:51

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