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

[ramr0d]

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

 

[haruspex]

-6500(Vdi) + 50(3042)

Momentum is a vector. How do you add vectors?

 

[ramr0d]

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

 

[SammyS]

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 ?

 

[haruspex]

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

Yes.