Inelastic Collision with Friction in car accident

[Abarak]

Homework Statement

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1400 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.25 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.650.

What was the speed (v) of car A (mph) just before the collision?

By how many mph was car A exceeding the speed limit 35.0 mph?

Homework Equations

I have tried
1.) m_1 v_1i + m_2 v_2i = (m_1+m_2)v_f (does not work)
2.) v_f^2-v_o^2=2-U_k(g)D (does not work)

The Attempt at a Solution

The first equation above I know will not work because it was designed for a perfect inelastic collision. I need to take into account the friction, that is why I used equation two.

v_f^2-v_o^2=2-U_k(g)D
v_f = 0 m/s
v_0 = ? (need to find for problem #1)
U_k = 0.650
g = 9.8 m/s^2
D = 7.25m

Any ideas to what I could be doing wrong?

 

[Doc Al]

The first equation above I know will not work because it was designed for a perfect inelastic collision. I need to take into account the friction, that is why I used equation two.

You need both of those equations. First there is an inelastic collision (equation #1), then there is sliding (equation #2).

Work backwards!

v_f^2-v_o^2=2-U_k(g)D
v_f = 0 m/s
v_0 = ? (need to find for problem #1)
U_k = 0.650
g = 9.8 m/s^2
D = 7.25m

Good. Solve for v_0, which is the speed of the stuck cars after the collision.

 

[Abarak]

Hmmm... now I understand. v_0 in problem 2 = v_f in problem 1.

Problem 2:
v_f^2-v_o^2=2-U_k(g)D
v_0 = 6.647 m/s

Problem 1:
m_1 v_1i + m_2 v_2i = (m_1+m_2)v_f
1400 v_1i + 2100*0 = (1400+2100)6.647
v_1i = 16.618 m/s

is 16.618 m/s right?

 

[Doc Al]

Hmmm... now I understand. v_0 in problem 2 = v_f in problem 1.

Exactly.

Problem 2:
v_f^2-v_o^2=2-U_k(g)D
v_0 = 6.647 m/s

Correct approach, but revisit your calculation. (I think you forgot to multiply by 2.)

Problem 1:
m_1 v_1i + m_2 v_2i = (m_1+m_2)v_f
1400 v_1i + 2100*0 = (1400+2100)6.647
v_1i = 16.618 m/s

Also the correct approach. But redo with revised numbers.

And when you're done you'll have to convert from m/s to mph in order to answer the questions.

 

[Abarak]

First I wanted to just say thanks for the help!

When I perform the calculation for v_f^2-v_o^2=2-U_k(g)D I always get 6.647 m/s ??

v_f = 0 m/s
v_0 = ?
U_k = 0.650
g = 9.8 m/s^2
D = 7.25m

0^2-v_o^2=2-0.650*9.8*7.25 = 6.647 m/s ?

(I think you forgot to multiply by 2.)

Where do I multiply by 2?

Also, thanks for the heads up on the unit conversion. I almost forgot.

 

[Doc Al]

Where do I multiply by 2?

Right here:

When I perform the calculation for v_f^2-v_o^2=2-U_k(g)D I always get 6.647 m/s ??

Spot the 2!
Let me rewrite that equation more clearly:
v_f^2-v_o^2= -2 \mu_k g D

 

[Abarak]

Thanks for the help Doc Al!

I was able to get the right answer of 53.748 mph for question 1 and 18.8 mph for question 2.

 

[kahwei]

Thx this hlp me solved similar question.But may i ask why we should consider the acceleration as 9.8 m/s^2 ?

 

[Doc Al]

But may i ask why we should consider the acceleration as 9.8 m/s^2 ?

Who says the acceleration is 9.8 m/s^2? (That's g, a constant describing the Earth's gravity.)

 

[Gliese123]

Thx this hlp me solved similar question.But may i ask why we should consider the acceleration as 9.8 m/s^2 ?

Because the average gravity of Earth is approximately 9.8 m/s², (9.80665). Oh well, the vertical acceleration. Therefore, it's might be considered as convenient. What do I know. ^_^
http://en.wikipedia.org/wiki/Gravity_of_Earth