Given:
Mass of ball A, mA = 1.3 kg
Initial Velocity of ball A, uA = 2 m/s to the right
Mass of ball B, mB = 0.9 kg
Initial velocity of ball B, uB = 0 m/s (at rest)
Final velocity of ball A, vA = 1.5 m/s
θA = 30 degrees above the horizontal
θB = 60 degrees below the horizontal.
Let's find the final speed of ball B, vB, after the collision.
This is an elastic collision.
Let's apply the Conservation of Momentum:
[tex]m_Au_A+m_Bu_B=m_Av_Acos_A+m_Bv_Bcos_B[/tex]
Thus, we have:
[tex]\begin{gathered} (1.3)(2)+(0.9)(0)=(1.3)(1.5)(cos30)+(0.9)v_B(cos60) \\ \\ 2.6+0=1.69+0.9v_Bcos60 \end{gathered}[/tex]
Now, we have:
[tex]\begin{gathered} 0.9v_Bcos60=2.6-1.69 \\ \\ v_Bcos60=\frac{0.91}{0.9} \\ \\ v_Bcos60=1.01 \\ \\ v_B=\frac{1.01}{cos60} \\ \\ v_B=2.02\text{ m/s} \end{gathered}[/tex]
Therefore, the final speed of ball B after the collision is 2.02 m/s.
• AInitiER:
2.02 m/s
