Combination of Speeds? Car Crash problem?

  • Thread starter ilovemynny
  • Start date
  • #1
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Homework Statement


Mr. Smith was traveling northbound on Ashland Blvd when his car broadsided Mr. Green’s car, as it traveled east on Lime Rd. When the cars collided, they stuck together and slid. Analysis of the skid marks before the cars collided, indicated that Mr. Green did not apply his brakes, but that Mr. Smith did. Examination of the skid marks indicates that the speed of Mr. Smith’s car was reduced by 2.3 m/s before hitting Mr. Greene’s car.

The two drivers were interviewed. Mr. Smith stated that he had a green light when he went through the intersection, and went through the intersection at the posted speed limit (16 m/s). Mr. Green, however contends that he had stopped for a red light, and did not proceed into the intersection until after it had turned green. Because he started from a stop, Mr. Green does not know what his actual speed was when the accident occurred.

Analysis of evidence at the scene revealed the following information about the movement of the cars after they collided.

Angle of movement: 39° north of east
Displacement of vehicles: 0.81 m

The investigating officer at the scene has recorded the following additional information:

Mr. Smith

Automobile: Ford Escort
Weight: 1556 kg

Mr. Green

Automobile: Ford Escort
Weight: 1556 kg

1. What combination of speeds for Mr. Smith and Mr. Green is most consistent with the displacement and angle of the vehicles after the collision?
2. Does the evidence at the scene tend to confirm or deny the stories given to the investigating officer by Mr. Smith and Mr. Green?
3. While it is not possible to determine which driver ran the red light, is there any evidence that would indicate that a ticket should be issued to either one or both of the drivers?

As part of the one the scene investigation, the coefficient of sliding friction between the tires and the road was found to be 0.45. .



Homework Equations


P tot Before: (m1)(v1) + (m2)(v2)
P tot After: (m1)(v1) + (m2)(v2)

On the Y Axis
(m1)(v1)(sin) = (m2)(v2)(sin)
On the X Axis
(m1)(v1)(cos) + (m2)(v2)(cos) = (m1)(vf)

Friction Force = Coefficient Friction x Normal Force

Momentum = m x v

Force = m x change in velocity divided by time
or Force x Time = m x change in velocity

The Attempt at a Solution


so what I thought I should do was use this formula:
P tot Before: (m1)(v1) + (m2)(v2)
P tot After: (m1)(v1) + (m2)(v2)
but i already know the final speed so this formula wouldn't work.. right??? or am i wrong???

So then I thought maybe I would have to use this formula:
On the Y Axis
(m1)(v1)(sin 32) = (m2)(v2)(sin)

On the X Axis
(m1)(v1)(cos 32) + (m2)(v2)(cos 58) = (m1)(vf)

but I know this problem is inelastic since the cars stick together and this formula if for elastic collisions.

I really don't know what to do and I have no idea why I need the coefficient of friction, can someone help me?
 

Answers and Replies

  • #2
56
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Let [tex]m[/tex] be the mass of the car (both cars have same masses) and [tex]v_{1}[/tex] and [tex]v_{2}[/tex] be the speeds at the moment of the crash of Mr. Smith's and Mr. Green's car respectively. Conservation of momentum always applies. In this particular case, you have

[tex]m( \vec{v_{1}} + \vec{v_{2}} )=2m\vec{v},[/tex]

where [tex]v[/tex] is the speed of the cars after they collide.

For displacement [tex]r[/tex], you have

[tex]\vec{r}=\vec{v}t+\frac{1}{2}\vec{a}t^{2},[/tex]

where [tex]a[/tex] is acceleration of the cars due to friction (that's where you need coefficient of friction [tex]\mu[/tex]).

[tex]-2m\vec{a}=2m\frac{\vec{v}}{t}=2\mu mg \frac{\vec{r}}{r}[/tex]

All these equations involve vectors. If you decompose them into components, you should be able to calculate everything you need.
 

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