Given:
• Initial velocity of Puck A, u1 = 5 m/s
,• Final velocity of Puck A, v1 = 2.0 m/s
,• θ = 30 degrees
,• Initial velocity of puck B = 0 m/s
Let's find the speed and direction of puck B after the collision.
Apply the Law of Conservation of momentum:
[tex]p_{1A}+p_{1B}=p_{2A}+p_{2B}[/tex]Thus, we have:
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bv_B[/tex]Since the two Pucks are identical, this means they have the same masses.
Now, we have the x- and y-components
[tex]\begin{gathered} x-component: \\ p_{1A}cos30+p_{2A}cos60=p_{1B}+0 \\ \\ y-component: \\ p_{1A}sin30-p_{2A}sin60=0+0 \\ p_{1A}sin30=p_{2A}sin60 \\ p_{1A}=\frac{p_{2A}sin60}{sin30} \end{gathered}[/tex][tex]\begin{gathered} p_{1B}=\frac{p_{2A}sin60}{sin30}cos30+p_{2A}cos60 \\ \\ p_{1B}=p_{2A}(\frac{sin60}{sin30}cos30)+cos60) \end{gathered}[/tex][tex].[/tex]Now, to find the speed of puck B after collision, we have:
[tex]\begin{gathered} v_{2B}=\sqrt{(v_{AB}cos\theta_1-u_A)^2+(v_Asin\theta)^2} \\ \\ v_{2B}=\sqrt{(2.0cos30-5)^2+(2.0sin30)^2} \\ \\ v_{2B}=3.42\text{ m/s} \end{gathered}[/tex]The speed of puck B after the collision is 3.42 m/s.
Since the puck has a speed after collision, this is an elastic collision.
To find the direction, we have:
[tex]\begin{gathered} \theta_1-\theta_2=90 \\ \\ \theta_2=\theta_1-90=30-90=-60^o \end{gathered}[/tex]Therefore, the direction is 60 degrees in the opposite direction.
ANSWER:
Speed of puck B after the collision = 3.42 m/s
Direction = 60 degrees in the opposite direction.
