0
$\begingroup$

The problems is as follows:

A 2000kg Ford was travelling south when it collided with your 1000kg sports car travelling west. The two badly-damaged cars became entangled in the collision and leave a skid mark that is 20 meters long in a direction 14◦ to the west of the original direction of travel of the Excursion. The wealthy Excursion driver hires a high-powered lawyer who accuses you of speeding throughthe intersection. Use your knowledge, plus the police officer’s report of the recoil direction, the skid length, and knowledge that the coefficient of sliding friction between the tires and road is $\mu=0.6$, to deduce the original velocities of both cars. Were either of the cars exceeding the 30mph speed limit?

My attempt: $$\sum \vec p_{initial}=2000v_1(-\hat x)+1000v_2(-\hat y)$$ $$\mu=\frac{F_f}{F_N},F_{N,(after collision)}=3000g,F_f=\mu g(3000)$$ So then using F=ma, $$\mu g(3000)=(3000)a,a=\mu g$$ $$d=\frac 12 at^2+v_0t$$ In this case, I know that d=20meters, and $a=\mu g$. However, I do not have a $v_0$. My thought was to solve this for t, and then ssee what the cars final velocity was. After the collision, I know that the momentum is: $$\sum \vec p_{after}=3000v_{after}(-\cos(14^\circ)\hat x-\sin(14^\circ) \hat y)$$ So I get lost on how to combine all of these equations to solve for the initial velocities of the two cars.

$\endgroup$
1
$\begingroup$
  1. Use the kinetic energy = work done by friction to find out the after impact velocity

    $$ \frac{1}{2} (m_{ford} + m_{sports}) v_{after}^2 = \mu (m_{ford} + m_{sports})\, g\, d $$

  2. Use conservation of linear momentum to solve for the initial velocities

    $$ \begin{aligned} m_{sports} v_{sports} & = (m_{ford} + m_{sports}) v_{after} \sin \theta \\ m_{ford} v_{ford} & = (m_{ford} + m_{sports}) v_{after} \cos \theta \end{aligned} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.