Is there a difference in the impact force in this two cases?

  • A car driving at 100km/h crashes a truck standing still on the road
  • The same car driving at 200km/h crashes the same truck from behind, but the truck was moving at 100km/h before impact

EDIT: Consider it happens in space in a frictionless road (as suggested below by sammy)


If the collisions took place in space, or on a frictionless road, they would be identical in effect. Whatever reference frame the collision is observed from, all observers would agree about the relative velocities before and after, and the amount of damage done, although they would disagree about absolute velocities.

Suppose the masses are $m$ and $M$ and the initial velocities $v$ and $0$ in case 1, and $2v$ and $v$ in case 2. Suppose also that in both cases the car and truck become attached by the collision, so they move as a single object afterwards. Applying conservation of momentum, the final velocities are $\frac{m}{m+M}v$ and $\frac{2m+M}{m+M}v$ respectively.

The amounts of KE dissipated are :

case 1 : $\frac12mv^2-\frac12(m+M)(\frac{m}{m+M})^2v^2=\frac12 \frac{mM}{m+M}v^2$

case 2 : $\frac12 m(2v)^2+\frac12 Mv^2 - \frac12 (m+M)(\frac{2m+M}{m+M})^2v^2=\frac12\frac{mM}{m+M}v^2 $

Exactly the same in both cases. (The same result - ie no difference in the amount of KE lost - occurs also if the vehicles move apart after the collision.)

If the collision takes place on a road which can supply friction, the only difference is that the vehicles can apply their brakes. If the stationary truck has its brakes on, it is more difficult to move during the collision, the collision time is shorter and the impact forces are greater. In this case the damage will be greater. If neither truck applies its brakes, the collision is the same in both cases. Both vehicles can keep moving after the collision, so not all of the kinetic energy is dissipated.


I am assuming that severity, means the force experienced by the bodies. Which in these cases are the change in momentum of the car.

Case 1: Car hits still truck

$$\begin{align}F_\textrm{car} &= m\times(100-0)\\ F_\textrm{car} &= 100~\textrm m\end{align}$$

Thus force on the truck will be $-~F_\textrm{car}$ by Newton's Third Law.

Case 2: Car hits moving truck

Assuming the car stops after hit,

$$\begin{align}F_\textrm{car} &= m\times(200-0)\\ F_\textrm{car} &= 200~\textrm m\end{align}$$

In this case, the truck change in momentum of the truck is not clear. But assuming that the truck gains a little momentum ($\delta(mv) \lt 100~\textrm m$), truck experiences less impact.

Summary: Case 1 - Equal damage to both Car and Truck, Case 2 - Severe damage to truck. You can also do a thought experiment of the same, it will make more sense.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.