Answer :
Answer:
2.333 m/s to the right
Explanation:
By the conservation of momentum, we can write the following equation:
[tex]\begin{gathered} p_i=p_f \\ m_1v_{1i}+m_2v_{i2}=m_1v_{1f}+m_2v_{2f} \end{gathered}[/tex]Where m1 and m2 are the masses of the balls, vi, and vf are the initial and final velocities. If we take to the right as a positive and to the left as negative, we get
m1 = 2 kg
v1i = 3 m/s
m2 = 3 kg
v2i = -1 m/s
v1f = - 2 m/s
v2f = ?
Replacing the values and solving for v2f, we get:
[tex]\begin{gathered} (2kg)(3m/s)+(3kg)(-1m/s)=(2kg)(-2m/s)+(3kg)v_{f2} \\ 6\text{ kg m/s}-3kg\text{ m/s = -4 }kg\text{ m/s + (3}kg)v_{f2} \\ 3kg\text{ m/s = -4 }kg\text{ m/s + (3}kg)v_{f2} \\ 3kg\text{ m/s + 4 }kg\text{ m/s = (3}kg)v_{f2} \\ 7kg\text{ m/s = (3}kg)v_{f2} \\ \frac{7\text{kg m/s}}{3kg}=v_{f2} \\ 2.333m/s=v_{f2} \end{gathered}[/tex]Therefore, the velocity of the 3 kg ball is 2.333 m/s to the right.