Question 8.6: In the collision pictured in Figure 8.10, two carts collide ...

Solution for (a)
As before, the equation for conservation of momentum in a two-object system is

$m_1 v_1 + m_2v_2 = m_1v^′_1 + m_2v^′_2 .$                  (8.53)

The only unknown in this equation is $v^′_2$ . Solving for $v^′_2$ and substituting known values into the previous equation yields

$v^′_2 = \frac{m_1 v_1 + m_2v_2 − m_1 v^′_1}{m_2}$                   (8.54)

= 3.70 m/s.

Solution for (b)
The internal kinetic energy before the collision is

$KE_{int} =\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$                  (8.55)

= 0.763 J.

After the collision, the internal kinetic energy is

$KE^′_{int} =\frac{1}{2} m_1 v_1^{′2} + \frac{1}{2} m_2 v_2^{′2}$                (8.57)

= 6.22 J.

The change in internal kinetic energy is thus

= 5.46 J.

Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.