# Finding Speed Before Crash?

## Homework Statement

The Known Facts of the Accident

[1] Ned was traveling North on Oak Avenue in his car of mass 2000kg
[2] Ed was traveling East on King Street in his truck of mass 3000 kg
[3] The two approached the intersection at the same moment.
[4] With his stoplight “green”, Ned had the right-of-way and entered the intersection.
[5] Ed admitted that he failed to stop for the light and entered the intersection at the same moment as Ned.
[6] The collision occurred in the middle of the intersection.
[7] Both cars locked together upon impact and slid to a stop.
[8] From the point of impact (determined by shattered headlights on the pavement), the cars traveled a distance of 6.0 meters along a line that measured 40o North of East.
[9] From data tables in their accident investigation manuals, the investigators determined that the coefficient of friction between the road and the sliding cars was 1.5

Questions to Be Answered

[A] Considering the road surface to be horizontal, what was the frictional force that acted upon the locked sliding cars to stop them?

What was the magnitude of the acceleration that this force produced on the sliding cars sliding together?

[C] Based upon the stopping distance from the point of impact and the acceleration, what was the speed of the two locked cars at the moment after the impact?

[D] What was the momentum of the two locked cars at the moment after the impact?

[E] What part of this momentum did Ed contribute? (this would be his momentum before impact)

[F] What part of this momentum did Ned contribute?

[G] What was Ed’s speed before the collision?

[H] What was Ned’s speed before collision?

The speed limit at the intersection was 35 mi/hr. The police will ticket drivers for speeding if they exceeded the speed limit by more than 5 mi/hr. Should either of the two drivers be ticketed for speeding? Defend your answer with calculations.

## The Attempt at a Solution

So these are the questions I've already got to:

My Work:

A) To find the friction force I used this formula: Ff = coefficient of friction x mass x 9.81
So my friction force came out to be 73575 N which turned into 70,000 N because of significant numbers.

B) F = m x a so 70000 = 5000a and a = 14 m/s/s

C) vf^2 = vi^2 + 2ad
vf^2 - vi^2 = 2ad
vf^2 = 2 x (14)(6)
vf^2 = 168
vf = 12.96
vf = 13 m/s

D) Momentum: P = m x v
P = (2000 + 3000) kg x (13 m/s)
P = 65000 kg m/s

E) Momentum of Ed: P = m x v
P = 3000 kg x 13 m/s
Ed: P = 39000 kg m/s

F) Momentum of Ned: P = m x v
P = 2000 kg x 13 m/s
Ned: P = 26000 kg m/s

I just don't know how to find their initial speed!?!?!?!
And I feel like I didn't do E and F right.

## Answers and Replies

collinsmark
Homework Helper
Gold Member
E) Momentum of Ed: P = m x v
P = 3000 kg x 13 m/s
Ed: P = 39000 kg m/s
Sorry, but you'll need to rethink the above. Remember, momentum is a vector, like velocity. But this 13 m/s figure, as you've expressed it above is a speed.

To determine the momentum in the East direction, you'll need to take that "40o North of East" statement into consideration somehow.

And the mass is not just Ed's car. You're calculating the North component of the total wrecked mass. So you'll need to get the mass of Ted's car back in there too.

Think of it another way. You already know the magnitude (which you've calculated) and direction (given in the problem) of the total momentum after the collision. Break this total momentum vector into its x and y components for parts E) and F).
F) Momentum of Ned: P = m x v
P = 2000 kg x 13 m/s
Ned: P = 26000 kg m/s
Same comments as above. You'll need to take the "40o North of East" statement into consideration to determined the North component of the momentum. Mass is the total mass, not just Ted's car.

I just don't know how to find their initial speed!?!?!?!
Once you have each car's respective momentum, you can use $\vec p = m \vec v$ to get each car's respective initial velocity.
[Edit: This is step where you can use the individual masses of the respective cars.]

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