Consider a collision between two pucks on a frictionless air hockey table. Puck 1 has a mass of 350 g and an initial velocity of 2.5 m/s in the +x direction. Puck 2 has a mass of 510 g and is initially at rest.
After the collision puck 1 has been deflected by 30 degrees and has a speed of 1.7 m/s. What is the speed and direction of puck 2 after the collision?
|Before|| p1ix = 0.35*2.5|
p1ix = 0.875 kg m/s
p2ix = 0
|p1iy = 0|
p2iy = 0
|After|| p1fx = 0.35*1.7*cos(30)|
p1fx= 0.515 kg m/s
p2fx = ?
|p1fy = 0.35*1.7*sin(30)|
p1fy = 0.2975 kg m/s
p2fy = ?
Appliyng conservation of momentum in the x-direction:
p1ix + p2ix = p1fx + p2fx
p2fx = p1ix - p1fx
p2fx = 0.875 - 0.515 = 0.36 kg m/s
Appliyng conservation of momentum in the y-direction:
p1iy + p2iy = p1fy + p2fy
p2fy = -p1fy = -0.2975 kg m/s
The total momentum of the second puck after the collision is found from:
p2f2 = p2fx2 + p2fy2
p2f = 0.467 kg m/s
The direction of puck 2 after the collision can be found from:
It's fine to just use the magnitudes - we know puck 2 has +x and -y components.
θ = 39.6 degrees