Answer:
Speed is 1.73 m/s and the angle is [tex]30^{o}[/tex] below the x-axis
Explanation:
From the law of conservation of linear momentum, this collision is considered elastic hence we apply the principle on both x-axis and y-axis
[tex]m_{a}*v_{ia} + m_{b}*v_{ib}= m_{a}*v_{fa} + m_{b}*v_{fb}[/tex] where [tex]m_{a}[/tex] and [tex]m_{b}[/tex] are masses of pucks A and B respectively, [tex]v_{ia}[/tex] and [tex]v_a{ib}[/tex] are initial velocities of pucks A and B respectively, and [tex]v_{fa}[/tex] and [tex]v_{fb}[/tex] are final velocities of pucks A and B respectively, From the law of conservation of linear momentum, this collision is considered elastic hence we apply the principle on both x-axis and y-axis
[tex]v_{bf}=v_{bf}cos\theta + v_{bf}sin\theta[/tex]
Since the pucks are identical, the masses are same hence the equation can be written as
[tex]v_{ia} + v_{ib}= v_{fa}+ v_{fb}[/tex]
The final velocity of puck A is found by
[tex]v_{fa}=1cos60(i) + 1sin60(j) [/tex]
The final velocity of puck B is found by
[tex]v_{fb}=v_{fb}cos\theta + v_{fb}sin\theta[/tex]
Since initial velocity of puck B is zero, the law of conservation of linear momentum along x axis will be
[tex] (2)i +0=1cos 60(\hat i) + v_{fb}cos\theta[/tex]
[tex]v_{fb}=1.5(\hat i) [/tex]
On y-axis
[tex]1sin 60(\hat j) + v_{fb}sin\theta=0[/tex]
[tex]v_{fb}=-0.86 (\hat j) [/tex]
Magnitude after collision= [tex]\sqrt (1.5^{2} + -0.86^{2})=1.73 m/s[/tex]
Direction=[tex]tan^{-1}=\frac {0.86}{1.5}=30^{o}[/tex]
The angle is [tex]30^{o}[/tex] below the x-axis and speed of puck B after collision is 1.73 m/s
