Question
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. (Figure 1) 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. Part A Let the speeds of drivers E and N prior to the collision be denoted by and , respectively. Find , the square of the speed of the two-car system the instant after the collision. Express your answer terms of and . = SubmitHintsMy AnswersGive UpReview Part Part B What is the kinetic energy of the two-car system immediately after the collision? Express your answer in terms of , , and . = SubmitHintsMy AnswersGive UpReview Part Part C 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. = SubmitHintsMy AnswersGive UpReview Part Part D Using the information given in the problem introduction and assuming that the third driver is telling the truth, determine whether driver N has reported his speed correctly. Specifically, if driver E had been traveling with a speed of exactly 12 meters per second before the collision, what must driver N's speed have been before the collision? Express your answer numerically, in meters per second, to the nearest integer. Take , the magnitude of the acceleration due to gravity, to be 9.81 meters per second per second. = m/s SubmitHintsMy AnswersGive UpReview Part Provide FeedbackContinue of 1
Explanation / Answer
Part A Make a momentum balance: m_e · v_e + m_n · v_n = (m_e + m_n) · v where v is the speed of the two cars instant after the collision. Because the cars have the same mass, this simplifies to v_e + v_n = 2v The square of the speed instant after collision is: v² = 1/4·(v_e + v_n)² Part B The kinetic energy of a translatory motion in general is given by: E = 1/2·m·v² Here: K = 1/8·(m_e + m_n)·(v_e + v_n)² Part C: The kinetic energy of the sliding cars is reduced by friction work until they come to rest. This work is given by: W = ? Fds from point of collision to the point where the cars stop. F is the frictional force: F = µ·(m_e + m_n) (µ is the friction coefficient) Hence : W = ? µ·(m_e + m_n)ds = µ·(m_e + m_n)·l ( l is the length of the skid marks) From K = W 1/8·(m_e + m_n)·(v_e + v_n)² = µ·(m_e + m_n)·l you can calculate the speed of driver N: v_n = v(8·µ·l) - v_e