0.51 m/s to the right.
Explanation
The two carts lock together. As a result, the collision is inelastic. Kinetic energy will not conserve. Still, momentum conserves.
What's the momentum p of the two carts?
Before the collisions:
- [tex]p_1 = m_1 \cdot v_1 = 0.180 \times 0.80 = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}[/tex];
- [tex]p_2 = m_2 \cdot v_2 = 0[/tex].
- Sum of momentum: [tex]p = p_1 + p_2 = 0.144\; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}[/tex].
Momentum conserves. As a result, [tex]p(\text{after collision}) = p(\text{before collision}) = 0.144 \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}[/tex]
Velocity is the same for the two carts after the collision. Let [tex]v[/tex] denote that velocity.
[tex]p(\text{after collision}) = m_1\cdot v+ m_2 \cdot v = (m_1 + m_2) \cdot v[/tex].
[tex]v = \dfrac{p(\text{after collision})}{m_1 + m_2} \\ \phantom{v} = \dfrac{0.144}{0.180 + 0.100} \\ \phantom{v} = 0.51 \; \text{m}\cdot \text{s}^{-1} [/tex].
Direction of the movement will stay the same. Both cars are now moving to the right.
