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Chapter 7: Problem 71

An automobile weighing \(13.6 \mathrm{kN}\) is moving at $17.0 \mathrm{m} / \mathrm{s}\( when it collides with a stopped car weighing \)9.0 \mathrm{kN} .$ If they lock bumpers and move off together, what is their speed just after the collision?

Short Answer

Expert verified
Answer: The velocity of both cars after the collision when they move together is approximately 9.95 m/s.

Step by step solution

01

1. Convert weights to masses

Given the weights in kN, we need to convert them to masses in kg. The conversion factor is 1 kN = 1000 N. Using the equation: $$ \mathrm{Weight} = \mathrm{Mass} \times \mathrm{Gravity}, $$ where gravity is approximately 9.81 \(\mathrm{m/s^2}\). We can convert the weights to masses as follows: $$ \mathrm{Mass}_{1} = \frac{13.6 \times 10^3 \mathrm{N}}{9.81 \mathrm{m/s^2}} \\ \mathrm{Mass}_{1} \approx 1386.34 \mathrm{kg} $$ $$ \mathrm{Mass}_{2} = \frac{9.0 \times 10^3 \mathrm{N}}{9.81 \mathrm{m/s^2}} \\ \mathrm{Mass}_{2} \approx 918.25 \mathrm{kg} $$
02

2. Calculate initial momenta

Recall that momentum is given by the product of mass and velocity: $$ \mathrm{Momentum} = \mathrm{Mass} \times \mathrm{Velocity} $$ We now calculate the initial momenta of the two cars before the collision: Car 1: $$ \mathrm{Momentum}_{1i} = \mathrm{Mass}_{1} \times \mathrm{Velocity}_{1i} = 1386.34 \mathrm{kg} \times 17.0 \mathrm{m/s} \approx 23565.78 \mathrm{kg \cdot m/s} $$ Car 2: $$ \mathrm{Momentum}_{2i} = \mathrm{Mass}_{2} \times \mathrm{Velocity}_{2i} = 918.25 \mathrm{kg} \times 0.0 \mathrm{m/s} = 0 \mathrm{kg \cdot m/s} $$
03

3. Apply conservation of linear momentum

The conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. Since the two cars lock bumpers and move together, it means they will share the same final velocity (\(\mathrm{V_f}\)). Using conservation of linear momentum: $$ \mathrm{Momentum}_{1i} + \mathrm{Momentum}_{2i} = (\mathrm{Mass}_{1} + \mathrm{Mass}_{2}) \times \mathrm{V_f} $$ Plugging in the values, we get: $$ 23565.78 \mathrm{kg \cdot m/s} + 0 \mathrm{kg \cdot m/s} = (1386.34 \mathrm{kg} + 918.25 \mathrm{kg}) \times \mathrm{V_f} $$
04

4. Solve for the final velocity

Now, we solve for the final velocity, \(\mathrm{V_f}\): $$ \mathrm{V_f} = \frac{23565.78 \mathrm{kg \cdot m/s}}{1386.34 \mathrm{kg} + 918.25 \mathrm{kg}} \\ \mathrm{V_f} \approx 9.95 \mathrm{m/s} $$ So, the speed of both cars just after the collision when they move together is approximately \(9.95 \mathrm{m/s}\).

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Most popular questions from this chapter

Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of \(0.50 \mathrm{m} / \mathrm{s},\) someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at $1.30 \mathrm{m} / \mathrm{s} .$ (a) What velocity does the other glider have? (b) Is the total kinetic energy of the two gliders after the collision greater than, less than, or equal to the total kinetic energy before the collision? If greater, where did the extra energy come from? If less, where did the "lost" energy go?
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