Question

Blocks A (mass 3.00 kg ) and B (mass 14.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A.Part A) Find the maximum energy stored in the spring bumpers.Part B) Find the velocity of block A when the energy stored in the spring bumpers is maximumPart C) Find the velocity of block B when the energy stored in the spring bumpers is maximum.Part D) Find the velocity of block A after the blocks have moved apart.Part E) Find the velocity of block B after the blocks have moved apart.

Answer 1

Answer:

Explanation:

This is a crash problem, the first thing we should observe is if the shock is elastic, in this case the total mechanical energy is conserved, that is our case since the blocks separate after the crash

Part A)

   As the mechanical energy is conserved we will write the energy before the crash and at the point of maximum compression (during the crash)

Initial   Em= K1+K2

            Em = ½ ma Voa² + ½ mb Vob²

End      Em = U          speed zero

          U= ½ ma Voa² + ½ mb Vob²

          U= ½  3  2² + ½ 14 (-0.5)²

          U = 7.75 J  

This is the maximum energy stored

Part B  and C

As the stored energy is maximum, the speed of the blocks is zero

      Va=  0 m/s

      Vb =  0 m/s

Part D and E

  For this part we will also use the conservation of the momentum of movement

Before the Shock     Po = ma Voa + mb Vob

After the shock         Pf = ma Vfa + mb Vfb

                 Po = Pf

        ma Voa + mb Vob = ma Vfa + mb Vfb

        3 2 + 14 (-0.5) = 3 Vfa + 14 Vfb

        -1 = 3 Vfa + 14 Vfb

As the shock is elastic, the mechanical energy is conserved let's write it in the same instants

Before Shock         Ko = ½ m to Voa² + ½ mb Vob²

After the crash       Kf  =  ½ m to Vfa² + ½ mb Vfb²

                            Ko = Kf

           ½ ma Voa² + ½ mb Vob² = ½ ma Voa² + ½ mb Vob²

           ½ 3 2² + ½ 14 0.5² = ½ 3 Vfa² + ½ 14 Vfb²

           7.75 = 1.5 Vfa² + 7Vfb²

We have an equation system of two equations and two unknowns that we can solve

        -1 = 3 Vfa + 14 Vfb

        7.75 = 1.5 Vfa² + 7Vfb²

          Vfa = (-1 - 14 Vfb) / 3

          Vfa² = (7.75 -7 Vfb²) /1.5

Equating the two equation and solving you can get Vbf

         (-1-14Vfb)²/9 = (7.75 -7 Vfb²) /1.5

      (1 + 28 Vfb + 196 Vfb²) = (7.75 - 7 Vfb²) 6

      196 Vfb² +28 Vfb +1 +42 Vfb² - 46.5 = 0

      196 Vfb² + 70 Vfb -45.5 = 0

We solve the second degree equation, the correct answer is that the speed decreases and the sign of the velocity of the body of greater mass does not change