- #31
brightesthalo
- 26
- 0
do i have to assume the distance of the suburu seeing as it hasn't been given
Suburu WRX and a Honda Accord have collided at an intesection
- Thread starter brightesthalo
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SUMMARY
A Subaru WRX and a Honda Accord collided at an intersection, with the Subaru traveling north and the Honda westbound about to turn right. Witness statements and physical evidence indicate that neither driver braked before the collision, and the Honda was stationary at a stop sign 6 meters from the intersection. The combined mass of the vehicles was 2600 kg, and they skidded 12.8 meters at an angle of N 7 degrees W. Calculations show that the Subaru was traveling at approximately 44.14 km/h, which is below the 80 km/h speed limit, and the visibility conditions suggest the Subaru was likely visible to the Honda driver before the turn.
PREREQUISITES- Understanding of basic physics concepts such as momentum and kinetic energy.
- Familiarity with the work-energy theorem and its application in collision analysis.
- Knowledge of vector components and trigonometry for analyzing motion.
- Ability to calculate forces, mass, and acceleration using Newton's laws.
- Study the work-energy theorem and its applications in collision scenarios.
- Learn about momentum conservation in two-dimensional collisions.
- Explore the principles of friction and its role in vehicle dynamics.
- Review vector decomposition techniques for analyzing motion in different directions.
Physics students, accident reconstruction specialists, and individuals interested in understanding vehicle dynamics and collision analysis.
- #32
andrevdh
- 2,126
- 116
The cars are mangled together and are sliding as a unit.
- #33
brightesthalo
- 26
- 0
still won't help me decide whether the suburu was going faster than 80 km/h would it?
- #34
andrevdh
- 2,126
- 116
The answer need to be appoached in several steps. This is one of them before you will be able to answer that question.
The combined wrecks start with some speed v and are decelerated to zero by the frictional force. What do you get their initial speed?
The combined wrecks start with some speed v and are decelerated to zero by the frictional force. What do you get their initial speed?
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- #35
brightesthalo
- 26
- 0
i would use the equation i stated above to find the combined wreck speed in the 12.8 m.
i tried that and got 12.26 m/s ---> 44.136km/h
that right?
i tried that and got 12.26 m/s ---> 44.136km/h
that right?
- #36
andrevdh
- 2,126
- 116
I get the same.
Now you can calculate the momentum of the combined wrecks after the collision. What will the direction and magnitude of this vector be?
Now you can calculate the momentum of the combined wrecks after the collision. What will the direction and magnitude of this vector be?
- #37
brightesthalo
- 26
- 0
so it would be p=mv
= 2600 x 12.26
= 31876
??
= 2600 x 12.26
= 31876
??
- #38
andrevdh
- 2,126
- 116
Units of the calculated momentum? What would the direction of the vector be?
- #39
brightesthalo
- 26
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? ok now I am lost here I am gussin 7 deg
- #40
andrevdh
- 2,126
- 116
Yes that is seven degrees west of north. The direction of the momentum vector is determined by the direction in which the object is moving. In this case the direction in which the wrecks are sliding.
What will the components of this momentum vector be in the north-south and east-west directions?
What will the components of this momentum vector be in the north-south and east-west directions?
- #41
brightesthalo
- 26
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would this by any chance be wher sin tan cos would be used? if not then i don't really know
- #42
andrevdh
- 2,126
- 116
Yes. The components of a vector is calculated with the angle that it makes with one of the chosen perpendicular directions, seven degrees, and the magnitude (size or length) of the vector, that is 31786 kgm/s in this case.
- #43
brightesthalo
- 26
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maybe i should sleep on it a bit its 1:10 am here so sorry for bothering u . ill try tacle this in the morning ( don't know how successful ill be)
- #44
brightesthalo
- 26
- 0
actually ill stay a little longer i really need this done
- #45
andrevdh
- 2,126
- 116
It is time for me to go home also. I might be online in about 16 hours time.
What you need to do is calculate the components of this vector and compare it with the momentum vectors before the collision. Since momentum is conserved the components need to be the same in both directions before and after collision.
What you need to do is calculate the components of this vector and compare it with the momentum vectors before the collision. Since momentum is conserved the components need to be the same in both directions before and after collision.
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- #46
brightesthalo
- 26
- 0
how did u work that out?
- #47
brightesthalo
- 26
- 0
ok that's ok itl be too late i have class tomorrow
thank u for your help anyhow
i appreciate it very much
thank u for your help anyhow
i appreciate it very much
- #48
andrevdh
- 2,126
- 116
You need to calculate the components of the momentum after the collision, p_a. These two components are such vectors that when added together will produce the vector p_a. They also form the sides of a right angle (its base and perpendicular) as indicated in the attachment, while the vector p_a are the hypotenuse of the triangle.
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