A spring is compressed between two cars on a frictionless airtrack. Car A has four times the mass of car B, MA = 4 MB, while the spring’s mass is negligible. Both cars are initially at rest. When the spring is released, it pushes them away from each other. Which of the following statements correctly describes the velocities, the momenta, and the kinetic energies of the two cars after the spring is released? Note: Velocities and momenta are given below as vectors. 1. ~vA = −~vB ~pA = −4~pB KA = 16 KB 2. ~vA = + 1 5 ~vB ~pA = + 4 5 ~pB KA = 4 25 KB 3. ~vA = + 1 4 ~vB ~pA = +~pB KA = 4 KB 4. ~vA = − 1 4 ~vB ~pA = −~pB KA = 1 4 KB 5. ~vA = −2~vB ~pA = −8~pB KA = 16 KB 6. ~vA = −~vB ~pA = −~pB KA = KB 7. ~vA = − 1 3 ~vB ~pA = − 2 3 ~pB KA = 4 3 KB 8. ~vA = − 1 2 ~vB ~pA = −2~pB KA = KB 9. ~vA = −4~vB ~pA = −16~pB KA = 64 KB 10. ~vA = − 1 4 ~vB ~pA = −~pB KA = 4 KB

Respuesta :

Answer:

10. True.     vA = - ¼ vB ,  pA = - pB

Explanation:

Let us propose the solution of this problem, as the cars are released let us use the conservation of the moment.

Initial before releasing the cars

         p₀ = 0

Final after releasing cars

        [tex]p_{f}[/tex] = mA vA + mB vB

        p₀ = [tex]p_{f}[/tex]

        0 = mA vA + mB vB

        vA = - mB / mA vB

         

They indicate that mA = 4 mB

        vA = - ¼ vB

Let's write the amount of movement for each body

        pA = mA vA = 4 mB (- ¼ vB

        pA = -mB vB

        pB = mB vB

        pA = - pB

Let's check the answers

1 False

2 False

3 False

4 false

5 False

6 False

7 False

8 False

9 False

10. True. The speed and amount of movement values ​​are correct