1.15 m/s to the left (3 sig. fig.).
Explanation
Momentum is conserved between the two balls if they are not in contact with any other object. In other words,
[tex]p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}[/tex]
[tex]m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}[/tex], where
- [tex]m[/tex] stands for mass and
- [tex]v[/tex] stands for velocity, which can take negative values.
Let the velocity of objects moving to the right be positive.
- [tex]m_\text{A} = 1.55\;\text{kg}[/tex],
- [tex]m_\text{B} = 0.752\;\text{kg}[/tex].
Before the two balls collide:
- [tex]v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}[/tex],
- [tex]v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}[/tex].
After the two balls collide:
- [tex]v_\text{A}[/tex] needs to be found,
- [tex]v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}[/tex].
Again,
[tex]m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}[/tex],
[tex]1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)[/tex].
[tex]v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}[/tex].
[tex]v_{\text{A,final}}[/tex] is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.
Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.
