- #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
- Start date
Click For Summary
Homework Help Overview
The discussion revolves around a collision between a Subaru WRX and a Honda Accord at an intersection, focusing on the dynamics of the accident, including the speeds of the vehicles involved, their masses, and the skid marks left after the collision. Participants are attempting to analyze the situation using principles of physics, particularly momentum and kinetic energy.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking
Approaches and Questions Raised
- Participants discuss using the work-kinetic energy theorem and momentum conservation to analyze the collision and the subsequent skidding. There are questions about the relevance of certain details, such as the distance of the Honda from the stop sign and its initial velocity. Some participants express uncertainty about how to incorporate all given information into their calculations.
Discussion Status
The discussion is ongoing, with participants sharing various approaches and calculations. Some have provided guidance on using the principles of physics to determine speeds and forces involved in the collision. Multiple interpretations of the problem are being explored, particularly regarding the initial conditions of the vehicles and the implications of the skid marks.
Contextual Notes
Participants note the ambiguity in the problem statement and the assumptions they must make, such as the initial velocity of the Honda and the conditions under which the collision occurred. There is also mention of the visibility of the vehicles and the implications of the skid marks for understanding the dynamics of the crash.
- #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?
Last edited:
- #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
- 0
? 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
- 0
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
- 0
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.
Last edited:
- #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.
Attachments
Similar threads
- · Replies 33 ·
- · Replies 1 ·