The script for an action movie calls for a small race car (of mass 1500 kg and l

Question

The script for an action movie calls for a small race car (of mass 1500 kg and length 2.8 m) to accelerate along a flat-top boat (of mass 3900 kg and length 15 m), from one end of the boat to the other, where the car will then jump the gap between the boat and a somewhat lower dock. You are the technical advisor for the movie. The boat will initially touch the dock, as in Figure 9-81; the boat can slide through the water without significant resistance; both the car and the boat can be approximated as uniform in their mass distribution. Determine what the width of the gap will be just as the car is about to make the jump.

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c09/fig09_81.gif

I've done this problem several times and I keep getting the wrong answer. I've tried 2.611 meters and 4.1667 meters. I used Xcom= m1x1+m2x2/(m1+m2). That finds the Xcom which is then set into a second Xcom equation that finds the value of x, which is added into both components of the x1 and x2 values of the equation.

Please provide explanations and any additional equations needed to solve this question.

Thanks!

Explanation / Answer

let us assume that ,   mass of the caris m1 and is of point mass which is initially (2.8 m/2 )=1.4 m from the  right end of the boat, mass of theleft end of the boat is m2 and is initially at x =0 where the  dock exists, and its left end is at 15m so the boats centre of mass is initially at x = 15m /2 =7.5 m   Hence, centre of mass of thesystem is ,   xcom = (m1x1 + m2 x2) /(m1 + m1)              =[(1500 kg) (15 m - 1.4 m) + (3900 kg) (7.5 m)] / (1500 kg + 3900kg)            =  49650 /   5400 = 9.194 m     the next case , when the car makes a jumpis near the left end of the boat which has moved from the shore an amount x   the value of centre of mass is still xcom now the carat this moment is assumed to be as a point mass 1.4 m from the left end so,    x'com = (m1x1 + m2 x2) /(m1 + m1)               =[(1500 kg) (x + 1.4 m) + (3900 kg) (7.5 m +x)] / (1500 kg + 3900 kg)           = xcom  = 9.194 m [(1500 kg) (x + 1.4 m) + (3900 kg) (7.5 m +x)] / (1500 kg + 3900 kg)                   = 9.194 m 1500 x + 1.4 x + 3900 x7.5 + 3900 x  = 49647.6     x = 3.776 m   thus, width of the gap will be just as the car is about to make the jump.

is = 3.776 m                          thus, width of the gap will be just as the car is about to make the jump.

is = 3.776 m                       

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