# How Do You Calculate the Impact Force in a Two-Car Collision?

• get_physical
In summary, the impact force on both cars can be found by analyzing the change in momentum of either car, using the equation F=m*dv/t. In this scenario, the calculations should be done for the 1200kg car, not both cars together. The force on the 1200kg car is 2200N, while the force on the 1000kg car is not relevant to this calculation.
get_physical

## Homework Statement

1200kg car moving at 9.72m/s runs into and sticks to a parked 1000kg car. The collision takes 0.15s and after the crash the cars wheels lock and they skid to a stop on u=0.40 road. What is the impact force on both cars?

J=F*t=m*dv

## The Attempt at a Solution

I've found the distance it takes for it to stop (3.68m) and also the speed after the crash (5.3m/s).

I used the momentum to divide by the time (0.15s) to get the impact force, but the answer is wrong.

Did you make sure to use the change in momentum? Otherwise, your calculations for the speed after the crash and distance may be off.

The calculations for the speed after the crash and the distance are correct. But I just can't find the impact force.

get_physical said:
The calculations for the speed after the crash and the distance are correct. But I just can't find the impact force.
Show your arithmetic. Maybe you are making a mistake.

What does it mean that the collision takes 0.15 seconds? That deformation takes 0.15 seconds after which the cars start skidding with the remaining energy?

Oh Ok, I think I understand. Think about kinetic energies. If you knew the amount of energy and the time it was being applied, you can find out the force affecting the target mass.
Before the collision, the "speeding" car as a kinetic energy of N. After the collision, the 2-car-mass has a kinetic energy of M. M is dissipated by friction over time. What happened to the N-M , though?

Last edited:
Dear get,

All this could have been short-circuited if you had shown why F = 1200 kg * (9.72 - 5.3)m/s / 0.15 s did not yield the right answer. Someone would have told you to check your calculation against 1000 kg * 5.3 m/s / 0.15 s

BvU said:
Dear get,

All this could have been short-circuited if you had shown why F = 1200 kg * (9.72 - 5.3)m/s / 0.15 s did not yield the right answer. Someone would have told you to check your calculation against 1000 kg * 5.3 m/s / 0.15 s

Thank you. I used F= 2200 (0 - 5.3m/s) /0.15
Why do we only use 1200kg car and why do we not use 1000kg car?

get_physical said:
Thank you. I used F= 2200 (0 - 5.3m/s) /0.15
Why do we only use 1200kg car and why do we not use 1000kg car?
To find the force on a car, you must analyze the change in momentum of that car. Not both together! You can use either car.

## 1. What is impact force after collision?

Impact force after collision is the force exerted on an object during a collision or impact. It is the result of the change in momentum of the object and is typically measured in Newtons (N).

## 2. How is impact force after collision calculated?

Impact force after collision can be calculated using the formula F = m * Δv, where F is the impact force, m is the mass of the object, and Δv is the change in velocity (speed and direction) of the object during the collision.

## 3. What factors affect impact force after collision?

The factors that affect impact force after collision include the mass and velocity of the objects involved in the collision, the type and duration of the impact, and the elasticity of the objects (i.e. how much they can deform and spring back during the collision).

## 4. How does impact force after collision relate to energy?

Impact force after collision is directly related to the amount of kinetic energy that is transferred during the collision. The greater the impact force, the more energy is transferred between the objects involved. This can result in objects breaking, deforming, or bouncing off each other.

## 5. Can impact force after collision be reduced?

Impact force after collision can be reduced by increasing the duration of the impact, increasing the elasticity of the objects, or decreasing the mass and/or velocity of the objects involved. This can be achieved through the use of safety features such as airbags, crumple zones, and seat belts in vehicles, or by designing objects to absorb and dissipate energy during a collision.

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