Question

Suppose the two carts have the same mass m. In the initial state, these two carts are moving toward each other with the same initial speed, vi, along a frictionless track (implying no net external forces acting on the two carts). These carts collide and the result is some final state. The three parts of this question are concerned with three different final states. A. Assume that the carts hit each other and stop (both carts are not moving). Draw a momentum chart for this situation: make a separate row for each cart. B) Assume that the carts bounce off each other so that the final state of the system has each cart moving oppositely to its initial motion but with the same speed. Draw a momentum chart for this situation. C) As in (B), assume that the carts bounce off each other with equal speeds and in opposite directions, but now assume that the final speeds are smaller than the initial speeds. Draw a momentum chart. D) For each case does the total momentum of the two cars change? How do the momentum charts tell you this? E) Is the total kinetic energy constant for all three cases? How do you know?

Answer 1

Answer:

in all three cases the total moment is zero

cases A and B the kinetic energy is conserved.  

In case C the velocity decreases so the kinetic energy decreases

Explanation:

This is a momentum conservation exercise

          p = mv

In order for the moment to be preserved, we must define a system formed by the two cars, so that the forces during coke have been internal.

Before crash

car 1      p₀₁ = m v₀

car 2     p₀₂ = - mv₀

pose us several situations, we analyze each one

A) After the crash the cars stop

      $p_{f}$ = 0

p₀₁      m v₀

p₀₂    -m v₀

p_{f}  0

B) After the collision, each vehicle reverses its direction

       

p₀₁            m v₀

p₀₂           -m v₀

p_{f1}       -m v₀

p_{f2}       m v₀

C) In this case some of the kinetic energy is lost which is converted into internal energy, for example, deformation, heat, friction.

Consequently the speed of the cars is

               v < v₀

p₀₁          m v₀

p₀₂         - m v₀

p_{f1}     -m v

p_{f2}    m v

D) in cases A and B the momentum is maintained, but in case C the total momentum is maintained, even when the speed of the cars decreases, this is pf_total = 0

In all cases the total impulse is zero

           p₀ = p₀₁ + p₀₂ = m v₀ - mv₀

           p₀_total = 0

in all three cases the total moment is zero

E) The total kinetic energy is the sum of the kinetic energy of each car

          K_total = K₀₁ + K₀₂

          K_total = ½ m v₀² + ½ m (-v₀)²

          K_total = m v₀²

we see that because it is squared, the sign of the velocity does not matter, therefore in cases A and B the kinetic energy is conserved.

In case C the velocity decreases so the kinetic energy decreases

         Kf_total < K₀_total

the missing energy is transformed into internal energy during sackcloth.

In the attachment we can see a vector diagram of the momentum in each case