Question

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?​

Answer 1

Speed is 1.73 m/s and the angle is
`30^(o)` 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

`m_(a)*v_(ia) + m_(b)*v_(ib)= m_(a)*v_(fa) + m_(b)*v_(fb)` where
`m_(a)` and
`m_(b)` are masses of pucks A and B respectively,
`v_(ia)` and
`v_a{ib}` are initial velocities of pucks A and B respectively, and
`v_(fa)` and
`v_(fb)` 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

`v_(bf)=v_(bf)cos\theta + v_(bf)sin\theta`

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

`v_(ia) + v_(ib)= v_(fa)+ v_(fb)`

The final velocity of puck A is found by

`v_(fa)=1cos60(i) + 1sin60(j)`

The final velocity of puck B is found by

`v_(fb)=v_(fb)cos\theta + v_(fb)sin\theta`

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

`(2)i +0=1cos 60(\hat i) + v_(fb)cos\theta`

`v_(fb)=1.5(\hat i)`

On y-axis

`1sin 60(\hat j) + v_(fb)sin\theta=0`

`v_(fb)=-0.86 (\hat j)`

Magnitude after collision=
`\sqrt (1.5^(2) + -0.86^(2))=1.73 m/s`

Direction=
`tan^(-1)=\frac {0.86}{1.5}=30^(o)`

The angle is
`30^(o)` below the x-axis and speed of puck B after collision is 1.73 m/s