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?

	X	Y
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:

p_1ix + p_2ix = p_1fx + p_2fx

p_2fx = p_1ix - p_1fx

p_2fx = 0.875 - 0.515 = 0.36 kg m/s

Appliyng conservation of momentum in the y-direction:

p_1iy + p_2iy = p_1fy + p_2fy

p_2fy = -p_1fy = -0.2975 kg m/s

The total momentum of the second puck after the collision is found from:

p_2f² = p_2fx² + p_2fy²

p_2f = 0.467 kg m/s

v2f

=

p2f m2

=

0.467 0.51

=

0.916 m/s

The direction of puck 2 after the collision can be found from:

tan(θ)

=

p2fy p2fx

It's fine to just use the magnitudes - we know puck 2 has +x and -y components.

θ = 39.6 degrees