Q104P The script for an action movie c... [FREE SOLUTION]

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Chapter 9: Q104P (page 255)

The script for an action movie calls for a small race car (of mass 1500 Kgand length 3.0 m ) to accelerate along a flattop boat (of mass 4000 kgand length 14 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 Fig. 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.

Short Answer

Expert verified

The width of the gap as the car is about to make the jump, d is 3.0 m .

Step by step solution

Understanding the given information

i) Mass of the car, $m_{1} i s 1500 k g$ .

ii) Length of the car, $d_{1} i s 3.0 m$ .

iii) Mass of the boat, $m_{2} i s 4000 k g$ .

iv) Length of the boat, $d_{2} i s 14 m$ .

Understanding the given information

You use the concept of center of mass. Using the given distances and masses, you can find the position of the center of mass. When there is no external force acting on the system, then the center of mass of the system remains at the same position. Using a reference point at the right end of the dock, you can find the gap between the dock and the boat, and the formula to find the above is given below

$x_{c o m} = \frac{(m_{1} x_{1} + m_{2} x_{2})}{(m_{1} + m_{2})}$

Calculation for the width of the gap just as the car is about to make the jump.

The Center of mass of the boat is initially at 7 m, and the center of mass of the car will be at its center, that is at 1.5 m.

You can find the center of mass of the boat鈥揷ar system as,

$x_{c o m} = \frac{(m_{1} x_{1} + m_{2} x_{2})}{(m_{1} + m_{2})}$ (1)

Let us consider the left end of the boat as the reference point, so the position of the center of mass of the car is $x_{1} = (d_{2} - 1.5)$ , and you get,

$x_{c o m} = \frac{1500 (14 - 1.5) + 4000 (7.0)}{(1500 + 4000)} x_{c o m} = 8.5 m$

When there is no external force acting on the system, there is no change in the center of mass. It will remain at the same point 8.5 m .

When the car is about to jump, it is at the left end of the boat. So, we can take the reference point from the right end of the dock, so the distance of the car from the reference point will be $x_{1} = (d + 1.5)$ .

Also, the distance of the center of mass of the boat will be $x^{2} = d + 7.0$ , where d is the gap between the dock and the boat. The center of mass of the system is 8.5 m .

We can write, and substitute these values in equation 1 as,

$8.5 = \frac{1500 (d + 1.5) + 4000 (d + 7.0)}{(1500 + 4000)} 8.5 脳 5500 = 1500 d + 2250 + 4000 d + 28000 d 46750 = 30250 + 5500 d = \frac{16500}{5500} = 3.0 m$