Question

In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart that is initially at rest. In trial 2, the setup is identical except the carts stick together during the collision. How does the speed of the two-cart system's center of mass change, if at all, during the collision in each trial

Answer 1

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          $x_{cm}$ = $\frac{1}{M} \sum x_i m_i$

in this case we have two bodies

          x_{cm} = $\frac{1}{M}$ (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = $\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )$

          x_{cm} = $\frac{1}{M} ( m_1 v_1 + m_2 v_2 )$

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = $\frac{m_1}{m_1+m_2}$  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         $v_{cm o}$ = $\frac{1}{m_1 +m_2}$ (m₁ vo)

final

         $v_{cm f}$ = $\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )$ ( v) = v

         v_{cm f} =  $\frac{m_1}{M^2} v_o$

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = $\frac{1}{M} = \frac{1}{m_1 +m_2}$

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = $\frac{m_1 -m_2}{m1 +m2 } v_o$

             v_{2f} = $\frac{2 m_1}{m-1+m_2} vo$

now let's find the velocity of the center of mass

initial

          $v_{cm o}$ = $\frac{1}{M}$ m₁ v₀

final

          $v_{cm f}$ = $\frac{1}{M}$  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = $\frac{1}{M}$ [  $m_1 \frac{m_2}{M}$ + $m_2 \frac{2 m_1}{M}$ ] v₀

          v_{cm f} = $\frac{1}{M^2}$ ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = $\frac{1}{M^2}$ (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = $\frac{1}{M}$ M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.