Question

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

Answer 1

After the spring between two cars on a frictionless track is released, car A (four times the mass of car B) will have a velocity that is -1/4 that of car B, momentum in magnitude equal but opposite direction, and 1/4 the kinetic energy of car B.

Step-by-step explanation:

In a problem where a spring is compressed between two cars of different masses on a frictionless track, when the spring is released, we must consider the principles of conservation of momentum and conservation of energy. Car A has four times the mass of car B, MA = 4 MB, and both cars start at rest. After the spring is released, car A will move with a velocity vA and car B will move with a velocity vB. By the conservation of momentum, the momentum of car A will be equal in magnitude and opposite in direction to the momentum of car B, and because MA = 4 MB, it follows that vA = -1/4 vB.

For their momenta, we write pA = MA * vA and pB = MB * vB. Therefore, pA = -pB, since MA = 4 MB and vA = -1/4 vB. Now, considering kinetic energy, KA = 1/2 MA vA^2 and KB = 1/2 MB vB^2, we find that KA = 1/4 KB because vA is 1/4 of vB, but car A has four times the mass of car B. Thus, the correct description of the cars' velocities, momenta, and kinetic energies is: vA = -1/4 vB, pA = -pB, KA = 1/4 KB, matching option 4 above.

Answer 2

10. True. vA = - ¼ vB , pA = - pB

Step-by-step 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

`p_(f)` = mA vA + mB vB

p₀ =
`p_(f)`

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