7. Two cars are moving. The first car has twice the mass of the second car, but only half as much kinetic energy. Then, both cars increase their speed by 8.00 m/s. They then have the same kinetic energy. The original speeds of the two cars are, respectively,

A. v1 = 5.66 m/s, v2 = 11.3 m/s.
B. v1 = 14.1 m/s, v2 = 7.07 m/s.
C. v1 = 11.3 m/s, v2 = 5.66 m/s.
D. v1 = 22.6 m/s, v2 = 11.3 m/s.
E. v1 = 14.1 m/s, v2 = 22.6 m/s.

À. v1 = 5.66 m/s, v2 = 11.3 m/s.

Let's solve this step by step:

1. Let's assume the initial masses of the first and second cars as m1 and m2, respectively.

2. Given that the first car has twice the mass of the second car, we can write:

m1 = 2m2

3. It is also given that the first car has half the kinetic energy of the second car initially, which means:

(1/2)mv1^2 = (1/2)mv2^2

Dividing both sides of the equation by (1/2)m, we get:

v1^2 = v2^2

4. Now, both cars increase their speed by 8.00 m/s. So, the final speeds of the cars will be:

v1_final = v1_initial + 8.00
v2_final = v2_initial + 8.00

5. It is also given that both cars have the same kinetic energy after the increase in speed. Therefore, we can write:

(1/2)m1v1^2 = (1/2)m2v2^2

Substituting the values of v1^2 and rearranging the equation, we get:

v2^2 = (2/1)v1^2

6. Now, let's substitute the values of v1_final and v2_final into the equation from step 5 and solve for v1_initial and v2_initial.

[(2/1)(v1_initial + 8.00)^2] = v2_initial^2

7. Rearranging the equation and simplifying, we get:

2(v1_initial^2 + 16v1_initial + 64) = v2_initial^2

8. Using the relationship m1 = 2m2 from step 2, we can substitute m1 = 2 and m2 = 1 into the equation from step 7:

2(v1_initial^2 + 16v1_initial + 64) = (2v1_initial)^2

9. Solving this equation, we get:

v1_initial = 5.66 m/s
v2_initial = 11.3 m/s

Therefore, the original speeds of the two cars are:
v1 = 5.66 m/s and v2 = 11.3 m/s

So, the correct answer is option A: v1 = 5.66 m/s, v2 = 11.3 m/s.

To solve this problem, we can use the formula for kinetic energy:

Kinetic energy (KE) = (1/2) * mass * velocity^2

Let's represent the mass of the second car as m2 and its original speed as v2.
Given that the mass of the first car is twice that of the second car, its mass can be represented as 2m2. Its original speed can be represented as v1.

From the given information, we know that the first car has half as much kinetic energy as the second car. Mathematically, this can be expressed as:

(1/2) * (2m2) * (v1^2) = (1/2) * m2 * (v2^2)

Simplifying this equation, we have:

v1^2 = (m2/m1) * v2^2 ...........(1)

We are also given that both cars increase their speed by 8.00 m/s, so we can rewrite the equation for the new kinetic energy as:

(1/2) * (2m2) * (v1 + 8)^2 = (1/2) * m2 * (v2 + 8)^2

Expanding and simplifying this equation, we get:

(v1 + 8)^2 = (v2 + 8)^2 ...........(2)

Now, we have a system of two equations (equation 1 and equation 2) with two unknowns (v1 and v2). We can solve this system of equations to find the values of v1 and v2.

Let's substitute equation 1 into equation 2:

((m2/m1) * v2^2 + 8)^2 = (v2 + 8)^2

Simplifying further, we get:

((m2/m1) * v2^2)^2 + 2 * ((m2/m1) * v2^2) * 8 + 8^2 = (v2 + 8)^2

Expanding and simplifying, we obtain:

(v2^2) * ((m2/m1)^2 - 1) = 0

Since (m2/m1)^2 is positive, the only possible solution is v2^2 = 0, which means v2 = 0.

Substituting this value of v2 into equation 1, we can solve for v1:

v1^2 = 2 * v2^2 = 0

Therefore, v1 = 0.

Since the original speeds of both cars are zero, none of the provided answer choices are correct for this problem.