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