# How Does Friction Impact Post-Collision Speed in Inelastic Car Accidents?

• ramr0d
In summary: Friction plays a role in the calculation of the Durango's speed just before the collision. The coefficient of kinetic friction between the ground and the wreckage is needed to determine the net force acting on the Durango and ultimately its speed.In summary, the Texas Department of Public Safety is investigating an accident involving a 2012 Prius and a 2013 Dodge Durango. The collision occurred in a remote area of the Texas Panhandle early on a foggy morning. The wreckage of the two vehicles was locked together and skidded across the ground until it struck a tree. The tree was 35 ft from the point of impact, in a direction 39∘north of east. The Prius was traveling at a speed of
ramr0d
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For the Texas Department of Public Safety, you are investigating an accident that occurred early on a foggy morning in a remote section of the Texas Panhandle. A 2012 Prius traveling due north collided in a highway intersection with a 2013 Dodge Durango that was traveling due east. After the collision, the wreckage of the two vehicles was locked together and skidded across the level ground until it struck a tree. You measure that the tree is 35 ft from the point of impact. The line from the point of impact to the tree is in a direction 39∘north of east. From experience, you estimate that the coefficient of kinetic friction between the ground and the wreckage is 0.45. Shortly before the collision, a highway patrolman with a radar gun measured the speed of the Prius to be 50 mphand, according to a witness, the Prius driver made no attempt to slow down. Four people with a total weight of 460 lb were in the Durango. The only person in the Prius was the 150-lb driver. The Durango with its passengers had a weight of6500 lb, and the Prius with its driver had a weight of 3042 lb.

What was the Durangos speed just before the collision?

How fast was the wreckage traveling just before it struck the tree?

I was able to find the first question by using conservation of momentum. Since one is going north and the other is going east, you have to set it up as components. First, I did the y component, and found the final velocity. I used trig to find the final velocity of the x component. You then use that final velocity and plug it into the x component of momentum to find the initial x velocity. This is the Durango's initial velocity.
I'm having trouble with the next question. I was able to find the velocity of the inelastic collision right after the collision, but I'm not sure where to from there.

Here's my math so far...
Md = mass of durango (kg)
Mp = mass of prius (kg)
d = 35 ft
theta= 39 degrees
Vp = velocity of prius (mph)
Vd = velocity of durango (mph)

x component mom: 6500*(Vdi)+0=9542(Vdf)
y component mom: = 50(3042) + 0 = 9542(Vpf)
Using trig,
tan(theta)= Vpf/Vdf
Solve for Vdf and plug back into x component momentum.

For the second part, I know that momentum is conserved in an inelastic collision, so I need Vf.
I tried using Vf^2=Vi^2 + 2ad
Vi is found by adding the two components together and equating to the final momentum
-6500(Vdi) + 50(3042) = 9542(Vi)
d is 35 feet, given from the equation, and I'm not sure how to get a.

Thanks for the help

ramr0d said:
-6500(Vdi) + 50(3042)
Momentum is a vector. How do you add vectors?

sqrt((6500(vdi))^2+(50(3042))^2)) for momentum vector?

ramr0d said:
For the Texas Department of Public Safety, you are investigating an accident that occurred early on a foggy morning in a remote section of the Texas Panhandle. A 2012 Prius traveling due north collided in a highway intersection with a 2013 Dodge Durango that was traveling due east. After the collision, the wreckage of the two vehicles was locked together and skidded across the level ground until it struck a tree. You measure that the tree is 35 ft from the point of impact. The line from the point of impact to the tree is in a direction 39∘north of east. From experience, you estimate that the coefficient of kinetic friction between the ground and the wreckage is 0.45. Shortly before the collision, a highway patrolman with a radar gun measured the speed of the Prius to be 50 mphand, according to a witness, the Prius driver made no attempt to slow down. Four people with a total weight of 460 lb were in the Durango. The only person in the Prius was the 150-lb driver. The Durango with its passengers had a weight of6500 lb, and the Prius with its driver had a weight of 3042 lb.

What was the Durangos speed just before the collision?

How fast was the wreckage traveling just before it struck the tree?

I was able to find the first question by using conservation of momentum. Since one is going north and the other is going east, you have to set it up as components. First, I did the y component, and found the final velocity. I used trig to find the final velocity of the x component. You then use that final velocity and plug it into the x component of momentum to find the initial x velocity. This is the Durango's initial velocity.
I'm having trouble with the next question. I was able to find the velocity of the inelastic collision right after the collision, but I'm not sure where to from there.

Here's my math so far...
Md = mass of durango (kg)
Mp = mass of prius (kg)
d = 35 ft
theta= 39 degrees
Vp = velocity of prius (mph)
Vd = velocity of durango (mph)

x component mom: 6500*(Vdi)+0=9542(Vdf)
y component mom: = 50(3042) + 0 = 9542(Vpf)
Using trig,
tan(theta)= Vpf/Vdf
Solve for Vdf and plug back into x component momentum.

For the second part, I know that momentum is conserved in an inelastic collision, so I need Vf.
I tried using Vf^2=Vi^2 + 2ad
Vi is found by adding the two components together and equating to the final momentum
-6500(Vdi) + 50(3042) = 9542(Vi)
d is 35 feet, given from the equation, and I'm not sure how to get a.

Thanks for the help
What do you suppose friction has to do with this ?

ramr0d said:
sqrt((6500(vdi))^2+(50(3042))^2)) for momentum vector?
Yes.

## What is an inelastic car collision?

An inelastic car collision is a type of collision where the kinetic energy of the system is not conserved, meaning that some of the initial kinetic energy is converted into other forms of energy, such as heat or sound.

## What is the difference between an inelastic and elastic car collision?

In an elastic car collision, the kinetic energy of the system is conserved, meaning that the total kinetic energy before and after the collision is the same. In an inelastic car collision, some of the initial kinetic energy is converted into other forms of energy, resulting in a decrease in the total kinetic energy of the system.

## What factors can affect the degree of inelasticity in a car collision?

The degree of inelasticity in a car collision can be affected by factors such as the speed and mass of the cars involved, the angle of collision, and the materials and structures of the cars.

## Why are inelastic car collisions more dangerous than elastic collisions?

Inelastic car collisions are more dangerous because the decrease in kinetic energy results in a larger amount of force being applied to the cars and their occupants. This can lead to more severe damage and injuries.

## How can the inelasticity of a car collision be measured?

The inelasticity of a car collision can be measured by calculating the coefficient of restitution, which is the ratio of the final to initial relative velocity of the two cars. A lower coefficient of restitution indicates a more inelastic collision.

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