Two identical pucks are on an air table. puck a has an
initial velocity of 2.0 m/s in the +x-direction. puck b is at
rest. puck a collides elastically with puck b and a moves
off at 1.0 m/s at an angle of 60° above the x-axis. what is
the speed and direction of puck b after the collision?

Speed is 1.73 m/s and the angle is 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  

where and are masses of pucks A and B respectively, and are initial velocities of pucks A and B respectively, and and 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  

Since the pucks are identical, the masses are same hence the equation can be written as  

The final velocity of puck A is found by  

The final velocity of puck B is found by  

Since initial velocity of puck B is zero, the law of conservation of linear momentum along x axis will be  

On y-axis  

Magnitude after collision=  

Direction=  

The angle is below the x-axis and speed of puck B after collision is 1.73 m/s  

mistakes are ok

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this