Car Crash and Such: Investigation and Analysis

[~christina~]

[SOLVED] Car Crash and such

Homework Statement

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

(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.

Homework 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
$$Kf= Ki - f_k d$$

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 separate 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!

 

[Astronuc]

but since it is at 90 deg before...after would the components of velocity have a angle?

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.

 

[~christina~]

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

 

[~christina~]

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

 

[Astronuc]

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.

 

[Bill Foster]

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

Use this to find initial speed:

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

 

[Bill Foster]

Vehicle 1 was traveling 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 traveling 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

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

 

[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

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

 

[~christina~]

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

Use this to find initial speed:

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

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


help...

 

[Doc Al]

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

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.

 

[~christina~]

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.

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

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

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

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

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

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

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

$$v_o= 9.29m/s$$


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

 

[Doc Al]

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

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

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

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

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

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

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

$$v_o= 9.29m/s$$


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

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.

 

[~christina~]

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.

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.

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

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)

 

[Doc Al]

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.)

 

[Doc Al]

Hm okay so they say that the angle they hit at is 90 degree angles...Not sure however about the angle after the collision and how I find that.

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 think it would be ...for momentum

$$m_1v_{1f} = m_1v_{1i}sin\theta +m_2v_{2i}sin\theta$$

$$m_2v_{2f} = m_1v_{1i}cos\theta + m_2v_{2i}cos\theta$$

Is this fine?

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)

I would think it would be however I don't have the initial v of both before the crash so how wouldn't I have 2 unknowns and an unknown angle too? Unless there is a way to find the angle after crash of a 90 degree collison.

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!

 

[~christina~]

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.

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!

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

well I thought I had to separate 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.

 

[Doc Al]

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

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

That's why I wrote it in vector form. You can separately equate each component of momentum. For example (the y component):
$$m_1v_1 = (m_1 + m_2)v_f\sin\theta$$

okay well I think that it's 45 deg for the angle after the collision

Sounds good to me.

but the problem is that they don't stick together the whole time and the picture shows them with 2 different angles.

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.

 

[~christina~]

That's why I wrote it in vector form. You can separately equate each component of momentum. For example (the y component):
$$m_1v_1 = (m_1 + m_2)v_f\sin\theta$$

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

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.

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

 

[Doc Al]

well I get.

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

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

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

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

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

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

Looks good.

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

That's the idea.

 

[Bill Foster]

Impulse is change in momentum.

 

[~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.

 

[Doc Al]

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.

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:
$$I_1 = m_1\vec{v}_f - m_1\vec{v}_i$$

 

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

[tex]1/2(934.40kg + 2,630.841kg)(9.29m/s)^2

$$KE_{after}= 15,3847.4579J$$

$$\Delta KE= KE_i - KE_f$$

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

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

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?

Not sure about this...do I do what I did in part a)?

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

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

 

[Doc Al]

change in momentum

$$\Delta p= mvf-mvi$$

impulse of car 1 when it hits car 2:

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

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.)

 

[~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

 

[~christina~]

wait...would it have a angle for the final velocity?

ex.

9.29m/s sin 45 => for the m1? since the velocity is not straight after but at an angle

 

[Doc Al]

hm..for the impulse I have a question...is it supposed to be negative?

It could certainly be negative--that just depends on how you've defined your coordinate system.

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

No good. In calculating $m_1\vec{v}_f - m_1\vec{v}_i$, realize that the initial and final velocities are not in the same direction. (Hint: Break them into components, subtract, then find the magnitude.)

according to this the m2 vehicle delivered a greater impulse

That should tell you that something is off. Since impulse equals $F\Delta t$, and Newton's 3rd law tells us that the forces that two bodies exert on each other must be equal (and opposite), you know that the impulse that they exert on each must also be equal and opposite. (Another, equivalent, way to look at it is to realize that momentum is conserved, so that whatever the change in momentum of one body, the other body must have the same (but opposite) change in momentum.)

wait...would it have a angle for the final velocity?

Oh yeah.

 

[~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

It could certainly be negative--that just depends on how you've defined your coordinate system.

No good. In calculating $m_1\vec{v}_f - m_1\vec{v}_i$, realize that the initial and final velocities are not in the same direction. (Hint: Break them into components, subtract, then find the magnitude.)

Wait..I'm not sure I get it however the reason is because I thought that the cars had momentums in ony 1 direction but thinking about it now gets me thinking that...
individual components of the final velocity would have a x and y component at the angle 45 deg, but I don't understand why it would be subtract to find the magnitude...


$$I_1= m_1v_f - m_2v_i$$

(934.40kg)(9.29m/s sin (45)) - (934.40kg)(25.06m/s)= -17277.9698 kg*m/s

$$I_2= m_2v_f-m_2v_i$$

(2630.84kg)(9.29m/s cos (45)) - (2630.84kg)(8.90m/s)= -6132.4301 kg*m/s

something is still off since they aren't equal and opposite in sign..

what am I doing incorrect?

___________________________

thought/Attempt #2:

since you say I need to find the magnitude of the velocity of the components then subtract them to find the magnitude I started plugging in numbers and got this.

(9.29m/s cos (45))- (9.29m/s sin (45))= 0

$$I_1= m_1v_f - m_2v_i$$

-0 - (934.40kg)(-25.06m/s)= -23,416.064 kg*m/s

$$I_2= m_2v_f-m_2v_i$$

-0 - (2630.84kg)(8.90m/s)= 23,414.476 kg*m/s

close but if it is correct it may be my rounding..."if" it's correct.

That should tell you that something is off. Since impulse equals $F\Delta t$, and Newton's 3rd law tells us that the forces that two bodies exert on each other must be equal (and opposite), you know that the impulse that they exert on each must also be equal and opposite. (Another, equivalent, way to look at it is to realize that momentum is conserved, so that whatever the change in momentum of one body, the other body must have the same (but opposite) change in momentum.)

oh...didn't relate those two..but now I'll remember that in the future.


Thanks

 

[Doc Al]

You must think of momentum as a vector.

Do this:

Find the x and y components of the initial momentum of car 1.

Find the x and y components of the final momentum of car 1.

Now find the x and y components of the change in momentum.

Then find the magnitude of the change in momentum.

 

[~christina~]

You must think of momentum as a vector.

Do this:

Find the x and y components of the initial momentum of car 1.

Find the x and y components of the final momentum of car 1.

Now find the x and y components of the change in momentum.

Then find the magnitude of the change in momentum.

starting with the car 1 momentums first

m1= 934.40kg
viy= 25.06m/s
vix= 0m/s

vfy= 9.29m/s sin 45
vfx= 9.29m/s cos 45

pxi= (934.40kg)(0m/s)= 0kg*m/s
pyi= (934.40kg)(25.06m/s)= 23,416.064kg*m/s

pxf= (934.40kg)(9.29m/s cos (45))= 61,389.09kg*m/s
pyf= (934.40kg)(9.29m/s sin (45)) = 6,1389.09kg*m/s

$$\Delta p_x= p_{xf}-p_{xi}= 61,389.09kg*m/s - 0kg*m/s = 61,389.09kg*m/s$$

$$\Deltap_y= p_{yf}-p_{yi}= 61,389.09kg*m/s - 23,416.064kg*m/s= 37,973.026kg*m/s$$

magnitude $$I= \Delta P = \sqrt{(61,389.09kg*m/s)^2 + (37,973.026)^2}= 72,184.285kg*m/s$$

If it is correct then I would think that the impulse is the impulse the car 1 delivered to car 2 when they crashed

Is this alright? (I hope I understood what you said)

However about the negative ...not sure where to put that...I was going to say that the momentum for the initial and final was negative since the car 1 goes in the y direction initially and finally goes in the y direction as well.

 

[Doc Al]

You seem to have the right idea now, but made an error here:

pxf= (934.40kg)(9.29m/s cos (45))= 61,389.09kg*m/s
pyf= (934.40kg)(9.29m/s sin (45)) = 6,1389.09kg*m/s

These should be: 6,138.09 kg-m/s
...

However about the negative ...not sure where to put that...I was going to say that the momentum for the initial and final was negative since the car 1 goes in the y direction initially and finally goes in the y direction as well.

Don't worry about the direction or sign of the impulse (although you could certainly find the direction if you wanted to, just like you'd find the direction of any other vector), just figure out the magnitude.

 

[~christina~]

You seem to have the right idea now, but made an error here:

These should be: 6,138.09 kg-m/s

Oh my... thanks for catching that Doc Al

well editing that...

m1= 934.40kg
viy= 25.06m/s
vix= 0m/s

vfy= 9.29m/s sin 45
vfx= 9.29m/s cos 45

pxi= (934.40kg)(0m/s)= 0kg*m/s
pyi= (934.40kg)(25.06m/s)= 23,416.064kg*m/s

pxf= (934.40kg)(9.29m/s cos (45))= 6,138.09kg*m/s
pyf= (934.40kg)(9.29m/s sin (45)) = 6,138.09kg*m/s

$$\Delta p_x= p_{xf}-p_{xi}= 6,138.09kg*m/s - 0kg*m/s = 6,138.09kg*m/s$$

$$\Deltap_y= p_{yf}-p_{yi}= 6,138.09kg*m/s - 23,416.064kg*m/s= -17,277.974kg*m/s$$

magnitude $$I= \Delta P = \sqrt{(6,138.09kg*m/s)^2 + (-17,277.974kg*m/s)^2}= 18,335.88kg*m/s$$

If it is now correct then I would think that the impulse is the impulse the car 1 delivered to car 2 when they crashed

Thanks

 

[Doc Al]

Looks good.

If it is now correct then I would think that the impulse is the impulse the car 1 delivered to car 2 when they crashed

Since you calculated the change in momentum of car 1, you found the impulse delivered to car 1. (Which was delivered by car 2, of course.)

 

[~christina~]

Looks good.

Since you calculated the change in momentum of car 1, you found the impulse delivered to car 1. (Which was delivered by car 2, of course.)

Oh..yep..


Thanks for all your help Doc Al