Part (A)
The total momentum of the system before collision is given as,
[tex]p_{ix}=m_1u_{1x}+m_2u_{2x}[/tex]Here, m_1 is the mass puck 1, m_2 is the mass of puck 2, u_1x is the x-component of initial velocity of puck 1 (u_1x=-10 m/s), and u_2x is the x-component of initial velocity of puck 2 (u_2x=8 m/s).
Substituting all known values,
[tex]\begin{gathered} p_{ix}=m\times(-10\text{ m/s})+m\times(8\text{ m/s}) \\ =-2m\text{ kg}.\text{ m/s} \end{gathered}[/tex]As, both puck does not have any initial velocity component along y-direction. The momentum along y-direction will be zero.
Therefore, the total momentum of the system before collision is,
[tex]\begin{gathered} p_i=p_{ix}\hat{i}+p_{iy}\hat{j} \\ =-2m\hat{i}\text{ kg}.\text{ m/s} \end{gathered}[/tex]Part (b)
According to the law of conservation of momentum, the momentum before and after the collision remains constant. Therefore, the total momentum after collision will be,
[tex]\begin{gathered} p_f=p_i \\ =-2m\hat{i}\text{ kg}.\text{ m/s} \end{gathered}[/tex]Part (C)
The x-component of the velocity of puck 1 after collision is given as,
[tex]v_{1x}=v_1\cos \theta[/tex]Here, v_1 is the velocity of puck 1 after collision (v_1=6 m/s), and θ is the angle made with x-axis (θ=25°).
Substituting all known values,
[tex]\begin{gathered} v_{1x}=(6\text{ m/s})\times\cos (25^{\circ}) \\ =5.44\text{ m/s} \end{gathered}[/tex]The y-componet of velocity of puck 1 after collision is given as,
[tex]v_{1y}=v_1\sin \theta[/tex]Substituting all known values,
[tex]\begin{gathered} v_{1y}=(6\text{ m/s})\times\sin (25^{\circ}) \\ =2.54\text{ m/s} \end{gathered}[/tex]Therefore, the velocity of puck 1 after collision in vector form is given as,
[tex]\begin{gathered} v_1=v_{1x}\hat{i}+v_{1y}\hat{j} \\ =(5.44\hat{i}+2.54\hat{j})\text{ m/s} \end{gathered}[/tex]Part (D)
Applying law of conservation of momentum along y-direction.
[tex]m_1u_{1y}+m_2u_{2y}=m_1v_{1y}+m_2v_{2y}[/tex]Here, v_2y is the velocity of puck 2 along y direction after collision.
Substituting all known values,
[tex]\begin{gathered} m\times0+m\times0=m\times(2.54\text{ m/s})+m\times v_{2y} \\ 0=2.54\text{ m/s}+v_{2y} \\ v_{2y}=-2.54\text{ m/s} \end{gathered}[/tex]Part (E)
Applying law of conservation of momentum along x-direction.
[tex]m_1u_{1x}+m_2u_{2x}=m_1v_{1x}+m_2v_{2x}[/tex]Here, v_2x is the velocity of puck 2 along x direction after collision.
Substituting all known values,
[tex]\begin{gathered} m\times(-10\text{ m/s})+m(8\text{ m/s})=m\times(5.44\text{ m/s})+mv_{2x} \\ -2m=5.44m+mv_{2x} \\ -2=5.44+v_{2x} \\ v_{2x}=-2-5.44 \\ v_{2x}=-7.44\text{ m/s} \end{gathered}[/tex]Part (F)
The angle at which puck 2 move after collision is given as,
[tex]\begin{gathered} \tan \phi=\frac{v_{2y}}{v_{2x}} \\ \phi=\tan ^{-1}(\frac{v_{2y}}{v_{2x}}) \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{-2.54\text{ m/s}}{-7.44\text{ m/s}}) \\ =18.85^{\circ} \end{gathered}[/tex]Part (G)
The magnitude of puck 2's velocity after the collision is given as,
[tex]v_2=\sqrt[]{v^2_{2x}+v^2_{2y}}[/tex]Substituting all known values,
[tex]\begin{gathered} v_2=\sqrt[]{(-7.44\text{ m/s})^2+(-2.54\text{ m/s})^2} \\ =7.86\text{ m/s} \end{gathered}[/tex]