# Car Collisons

by la673
Tags: collisons
 P: 13 plz help if u can. Consider the following two-car accident: Two cars of equal mass collide at an intersection. Driver E was traveling eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked wheels, before coming to rest. Police on the scene measure the length of the skid marks to be 9 meters. The coefficient of friction between the locked wheels and the road is equal to 0.9. i kno this question has been posted in other forums, but im tryin to work out the work doen by friction and keep gettin it wrong, Write an expression for the work done on the cars by friction. Express your answer symbolically in terms of the mass of a single car, the magnitude of the acceleration due to gravity , the coefficient of sliding friction , and the distance through which the two-car system slides before coming to rest. i get 2*mu*m*g*d which it (the web program) tells me im off by a multiplicative factor... Each driver claims that his speed was less than 14 meters per second (50 mph). A third driver, who was traveling closely behind driver E prior to the collision, supports driver E's claim by asserting that driver E's speed could not have been greater than 12 meters per second. Take the following steps to decide whether driver N's statement is consistent with the third driver's contention.
 P: 150 It's doing negative work on the car. Just add a negative sign in. BTW, do you go to sydney uni at all?
 P: 13 yup, so do u i take it? nd thx for that it was driving me crazy... couldnt work out where i was wrong
P: 150

## Car Collisons

No probs. The whole physics asignment is driving me crazy...
 P: 13 i kno wat u mean, now im stuck on the next question... got 12.61 m/s which is wrong... then i used 13 (cause it says to nearest integer) nd its also wrong
 P: 150 Do you have msn? Maybe we can chat through there.
 P: 13

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