# Homework Help: Car Crash and such

1. Nov 20, 2007

### ~christina~

[SOLVED] Car Crash and such

1. The problem statement, all variables and given/known data

At 1320 on the last Friday in September 1989 a frantic call was received at the local police station. There had been a serious automobile accident at the intersection of Main Street and State Street, with injuries involved. Lt. John Henry arrived at the scene 10 minutes after the phone call and found that two cars had collided at the intersection. In one car, the driver was unconscious and in the other car both driver and one passenger were injured.

A sketch of the accident scene is shown below. Main Street, a thoroughfare, has a 45 mile per hour speed limit. State Street also has a 45 mile per hour limit, but has a stop sign on either side of the road. Vehicle 2, which weighs 5800 lbs., skidded for 24 feet before coming to a stop next to the utility pole, marked Dec #20. Vehicle 1, which weighs 2060 lbs., showed no skid marks after the impact and came to a rest next to the house on the corner. Looking at the impact areas of the cars, it was clear to Lt. Henry that the cars impacted at right angles, hitting the front right bumper of vehicle 2 and the front left bumper of vehicle 1. After impact, they initially were traveling in the same direction. Lt. Henry noted that the weather was clear at 690, and that the roadway was dry. He used a drag sled to determine that the coefficient of friction between the tires and road was 0.60. He can't use the drag sled to determine the coefficient of friction between the tires of vehicle 1 as they roll over the roadway and grass, but he knows he can calculate that value.

John Henry has to use his reconstruction of the accident to determine whether the driver of vehicle 2 ran the stop sign and / or if the driver of vehicle 1 was speeding. Assume that the collision occurs in the middle of the intersection. (a) Find the velocities of the two vehicles just prior to impact. The mks system of units must be used for all calculations.

http://img229.imageshack.us/img229/4585/carcolidenm2.th.png [Broken]

(b) What is the total energy loss for the cars during the collision?

(c) What is the impulse delivered to car 1 when it is hit by car 2? ; the impulse delivered to car 2 by car 1?

(d) Which vehicle delivers the greater force of impact? Use the appropriate physics principle(s) to support your response.

(e) Lt. Henry measured the skid marks made by both vehicles prior to impact. The skid marks for vehicle 1 were 17 feet in length and for vehicle 2 were 10 feet in length. How fast was each car going just prior to braking?

(f) Which driver do you cite in the accident? Explain your response.

2. Relevant equations
$$\Delta momentum= P_f- P_i= mv_f- mv_i= I$$

$$m_1v_{1i} + m_2v_2i = m_1v_{1f} + m_2v_{2f}$$ ==> but since it is at 90 deg before...after would the components of velocity have a angle?

I'm not sure about energy though but I think simply would be
$$Kf= Ki - f_k d$$

3. The attempt at a solution

I think it is a Inelastic collision b/c they initially travel in same direction after collision but I'm not quite sure if it is since they don't really stick to each other but go their own seperate ways awhile after the collision.

well the info given:

Cars:
m1= 2060lbs=> 934.40 kg
d1= ?

m2= 5800lbs=> 2,630.84 kg
d2= 24ft=> 7.32m

impact angle= 90 deg

$$\mu_{k(between car and road)}= 0.60$$ ===> for which car, Not sure since it just says "car"

$$\mu_{k(between car and grass)}= ?$$ => they say I can find this but I'm not sure how to get it

a) velocities of cars just prior to impact

well it would be conservation of momentum...

after the collisions the angle would matter
which would be cos and which would be sin to find the angle I'm not sure about

They stick right after but not later after collison.. => is it inelastic?

I drew a orange line for the x and y axis after collision angles.

I'm thinking about momentum of system before and momentum of system after

but I don't know the initial velocities..

HELP!!

Last edited by a moderator: May 3, 2017
2. Nov 20, 2007

### Astronuc

Staff Emeritus
The combined momentum before the accident would have an angle, since both velocities are orthogonal (i.e. perpendicular to each other).

One car skids, but the other doesn't, so one decelerates possibly by interacting with the other car or meets resistance from grass/lawn or striking other objects.

The collision is inelastic, so KE is lost.

3. Nov 20, 2007

### ~christina~

but how would I find the velocities of the cars before impact if I don't have the final velocity of both cars??

4. Nov 20, 2007

### ~christina~

If anyone can help me I'd really appreciate it...Please?

5. Nov 21, 2007

### Astronuc

Staff Emeritus
The angle at which they both traveled a key part.

The fact that it is inelastic complicates the problem since one cannot use conservation of KE in addition to conservation of momentum before and after the collision, but one can use the distances traveled to determine the KE after the collision. The skidding car (2) is straightforward, but vehicle 1 rolls. However one could assume constant deceleration.

One must think as to whether the cars decelerate together. Are the traveling at the same angle in parallel? Determine the acceleration rate of 2 with just it's mass and how far it would travel without interaction of car 1, then add the mass of car 1 and see what the deceleration would be.

6. Nov 21, 2007

### Bill Foster

One vehicle weighs 5800 lbs, skidded for 24 feet with a friction coefficient of 0.60.

Use this to find initial speed:

$$v^2-v_0^2=2a\Delta{x}$$

7. Nov 21, 2007

### Bill Foster

Vehicle 1 was travelling in the y-direction, therefore it's momentum alone is responsible for the y-direction momentums of the cars after the collision.

Likewise...

vehicle 2 was travelling in the x-direction, therefore it's momentum alone is responsible for the x-direction momentums of the cars after the collision.

Therefore...

$$m_1v_1=m_1v_3sin(\theta)+m_2v_4sin(\theta)$$

and

$$m_2v_2=m_1v_3cos(\theta)+m_2v_4cos(\theta)$$

8. Nov 21, 2007

### Bill Foster

$$v_4$$ can be calculated...

$$v^2-v_0^2=2a\Delta{x}$$
$$v=\sqrt{2\mu g x}$$

So...

$$m_1v_1=m_1v_3sin(\theta)+m_2\sqrt{(2\mu g x)}sin(\theta)$$

and

$$m_2v_2=m_1v_3cos(\theta)+m_2\sqrt{(2\mu g x)}cos(\theta)$$

Last edited: Nov 21, 2007
9. Nov 25, 2007

### ~christina~

I don't get how I would know the final velocity
or acceleration of the m2 vehicle.

help...

10. Nov 25, 2007

### Staff: Mentor

The final velocity of m2 is zero: "skidded for 24 feet before coming to a stop next to the utility pole"

You can figure out the friction force on it, since you're given the coefficient of friction. Use that to figure out the acceleration. Then use the distance it skidded to figure out its initial speed.

11. Nov 26, 2007

### ~christina~

$$v^2 - v_o^2 = 2 a \Delta x$$

$$F_{fric} = \mu mg$$

$$F_{fric} = 0.60(2630.841kg)(9.81m/s^2)$$

$$F_{fric} = 15,485.13N$$

$$a= \frac{F_{net}} {m} = \frac{ 15,485.13N} {2,630.841kg} = -5.89m/s^2$$

$$v^2 - v_o^2 = 2 a \Delta x$$

$$o^2 - v_o^2 = 2(-5.98m/s^2)(7.32m)$$

$$v_o= 9.29m/s$$

$$vo_{m2}= 9.29m/s$$

Last edited: Nov 26, 2007
12. Nov 26, 2007

### Staff: Mentor

Looks good to me. (FYI you can generally save a bit of effort by sticking to symbols as long as possible. Since $F = \mu m g$, you can immediately deduce that $a = F/m = \mu g$ without needing to calculate F.)

If I understand the problem correctly, this speed you just calculated is the speed of both cars immediately after the collision. (Since they are moving together right after the collision.) Now use conservation of momentum to figure out the speeds of each car before the collision. In order to do that, you need the angle of the post-collision velocity. Either they give that to you somewhere or you're supposed to get it from the diagram.

13. Nov 26, 2007

### ~christina~

I thought I calculated the velocity before the collision of the vehicle m2

especially since you said that I used the distance the car skidded with the coeffient of friction is what I was going to use to find the initial v of the car.

thus..I found the initial v of m_2 right before collision right? or did I miss something?

because you have me confused now.

(I was confused about how to find the initial v of the m1 vehicle next)

14. Nov 27, 2007

### Staff: Mentor

Sorry if I wasn't clear. The 24 ft skid mark was made after the collision, so by "initial" speed I meant the speed at the start of the skid, which is the speed of the car just after the collision. (Pre-collision skid marks are discussed in part e.)

15. Nov 27, 2007

### Staff: Mentor

Since part of what you are trying to find are the speeds just prior to the collision, we need the angle the cars make after the collision. Either it's given or just make a guess based upon the diagram.
I look at it a bit differently.

The initial momentum (just prior to the collision) is:
$$m_1v_1\hat{j} + m_2v_2\hat{i}$$

The final momentum (just after the collision) is $(m_1 + m_2)v_f$, which you can write in terms of components as:
$$(m_1 + m_2)v_f\sin\theta \hat{j} + (m_1 + m_2)v_f\cos\theta \hat{i}$$

(where theta is the angle the final velocity makes with the x-axis)

You have the magnitude of the post-crash momentum (since you figured out the speed). If you knew the angle, you could figure out the precrash speeds. Estimate it!

16. Nov 27, 2007

### ~christina~

$$m_1v_1\hat{j} + m_2v_2\hat{i}$$= $$(m_1 + m_2)v_f\sin\theta \hat{j} + (m_1 + m_2)v_f\cos\theta \hat{i}$$

well I thought I had to seperate the x and y momentums...

okay well I think that it's 45 deg for the angle after the collision but the problem is that they don't stick together the whole time and the picture shows them with 2 different angles.

Last edited: Nov 27, 2007
17. Nov 27, 2007

### Staff: Mentor

That's why I wrote it in vector form. You can separately equate each component of momentum. For example (the y component):
$$m_1v_1 = (m_1 + m_2)v_f\sin\theta$$
Sounds good to me.
They move together for a while, that's all that counts. After that other factors enter the picture--such as the different friction forces on each car, one car is on the grass, etc. But we don't care about those issues.

18. Nov 27, 2007

### ~christina~

well I get.

$$m_1v_1\hat{j} = (m1 + m2)v_f sin \theta \hat {j}$$

$$(934.40kg)v_1\{j} = (934.40kg + 2,630.84kg)(9.29m/s)sin (45)$$

$$v_1\hat{j}= 25.06m/s$$

$$m_2v_2\hat{i} = (m1 + m2)v_f cos\theta \hat {i}$$

$$(2630.84kg)v_2\hat{i}= (934.40kg + 2,630.84kg)(9.29m/s) cos (45)$$

$$v_2\hat {i} = 8.90m/s$$

b) total energy loss for cars during collison

(I'm working on this now)

thinking of:

$$KE_{m1} + KE_{m2} =$$before collison

$$KE_{(m1+m2)}=$$after collision

oh..that was one of the main things that had me confused.

Last edited: Nov 27, 2007
19. Nov 27, 2007

### Staff: Mentor

Looks good.

That's the idea.

20. Nov 28, 2007

### Bill Foster

Impulse is change in momentum.

21. Nov 28, 2007

### ~christina~

I did the work but I shall post and see if it is correct after I hand in the assignment since I think it looked alright from my perspective.

Last edited: Nov 28, 2007
22. Nov 28, 2007

### Staff: Mentor

You deleted your last post before I had a chance to respond. Your calculation of KE change looked OK. But your thinking seemed a bit off regarding impulse. You need the change in momentum of each car separately. For example:
$$I_1 = m_1\vec{v}_f - m_1\vec{v}_i$$

23. Nov 28, 2007

### ~christina~

well here it is... I'll fix the impulse and repost

b) total energy loss for cars during collison

(I'm working on this now)

thinking of:

$$KE_{m1} + KE_{m2} =$$before collison

$$KE_{(m1+m2)}=$$after collision

before:$$1/2m_1v_1^2 + 1/2 m_2v_2^2$$

$$1/2(934.40kg)(25.06m/s)^2 + 1/2(2,630.84kg)(8.90m/s)^2$$

$$KE_{before}= 397,597.7001J$$

after: $$1/2 (m_1 +m_2) v_f^2$$

$$1/2(934.40kg + 2,630.841kg)(9.29m/s)^2 [tex] KE_{after}= 15,3847.4579J$$

$$\Delta KE= KE_i - KE_f$$

$$397,597.7001J- 153,847.4579J =$$

$$\Delta KE= 243,750.2422J$$

c.) Impulse delivered to car 1 when it is hit by car 2? Impulse delivered to car 2 by car1?

Hm..Impulse = change in momentum so I think I would use the momentum right after the crash and subtract that from the momentum right before the crash...thus

$$I= \Delta (mv)$$

and

for impulse delivered to car 2 by car 1 it would be...

$$vi_2\hat{i} = 8.90m/s$$

$$vi_1\hat{j} = 25.06m/s$$

$$vf= 9.29m/s$$

change in momentum

$$\Delta p= mvf-mvi$$

impulse of car 1 when it hits car 2:

$$(934.40kg + 2630.841kg)(9.29m/s) - (934.40kg)(25.06m/s) = 9705.02 kg*m/s$$

impulse of car 2 when it hits car 1:

$$(934.40kg + 2630.841kg)(9.29m/s) - (2630.841kg)(8.90m/s) = 9706.59 kg*m/s$$

d) which impulse is greater?

I say that the impulse that car 2 delivers to car one is greater that the impulse car 2 delivers to car 1.

e) skid marks for vehicle 1 were 17ft in length and for vehicle 2 were 10 ft in length.

HOw fast was each car going just prior to breaking. Which do you site in the accident?

I guess I just use the coeffient of friction as the one given for the road which is

$$\mu= 0.600$$

$$F_{fric}= \mu mg$$

I think however that I will use the speeds right before braking as the final velocity.

$$F_{fric}= \mu mg$$

vi=?
vf= 25.06m/s

$$Fm1_{fric}= 0.60(2630.841kg)(9.81m/s^2) = 15485.13N$$

Last edited: Nov 28, 2007
24. Nov 28, 2007

### Staff: Mentor

Two problems:
(1) Your first term is not the momentum of one car, but both cars.
(2) Initial and final velocities are in different directions. Momentum is a vector! (Subtract them like vectors, not just numbers.)

25. Nov 28, 2007

### ~christina~

hm..for the impulse I have a question...is it supposed to be negative?
I have an example in my text that has a elasic collison not inellastic collison and the velocity is postive b/c the negatives cancel out but when I calculated the momentum for m1 I got a negative:

$$I= \Delta(mv)= mv_f- mv_i$$

$$I_1= m_1v_f - m_2v_i$$

(934.40kg)(9.29m/s) - (934.40kg)(25.06m/s)= -14735.488 kg*m/s

$$I_2= m_2v_f-m_2v_i$$

(2630.84kg)(9.29m/s) - (2630.84kg)(8.90m/s)= 1026.03 kg*m/s

that is a large difference...did I do anything incorrect ?

according to this the m2 vehicle delivered a greater impulse