# Conservation of Momentum?

A hockey puck with mass 0.170 kg traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at 1.50 m/s strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is 30.0° away from the blue line at a speed of 0.750 m/s (see the figure). What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision? Relevance

Conservation of momentum along blue line

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.170)(1.5) + (0.170)(0) = (0.170)(0.750 cos30) + (0.170)(v₂cosθ)

0.255 = 0.110 + (0.170 v₂cosθ)

0.170 v₂cosθ = 0.145

v₂cosθ = 29/34 ....(1)

Conservation of momentum along perpendicular to blue line

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.170)(0) + (0.170)(0) = (0.170)(0.750 sin30) + (0.170)(v₂sinθ)

0 = 0.0638 + (0.170 v₂sinθ)

0.170 v₂sinθ = - 0.0638

v₂sinθ = -319/850 .....(2)

(2) divided by (1) gives

tanθ = (-319/850)/(29/34)

θ = -24°

plug in (1)

v₂cos(-24) = 29/34

v₂ = 0.934 m/s

The 2nd puck has magnitude of 0.934 m/s and direction of 24° on the other side of the blue line

KE conservation check !

KE before collision= ½m₁u₁² +½ m₂u₂² = ½m(u₁² + u₂²) = ½(0.170)(1.5² + 0²) = 0.191 J

KE after collision = ½(0.170)(0.750² + 0.934²) = 0.122 J

Since the KE was lost after collision, this is inelastic collision

• Let m be the common mass. m₁ = m₂ = 0.170 kg

Conservation of momentum in y direction

m₁(0) + m₂(0) = m₁(0.750sin30) + m₂(v₂sinθ)

v₂sinθ = -0.375 m/s

Conservation of momentum in x direction

m₁(1.50) + m₂(0) = m₁(0.750cos30) + m₂(v₂cosθ)

1.50 - 0.6495 = v₂cosθ = 0.85048

v₂sinθ / v₂cosθ  = -0.375 / 0.85048

tanθ = -0.44092...

θ = -23.7939... ≈ -23.8°  ANSWER

v₂sin-23.8 = -0.375

v₂ = 0.92948... ≈ 0.929 m/s  ANSWER

KEi = ½(0.170)(1.50²) + ½(0.170)(0²) = 0.19125 J

KEf = ½(0.170)(0.750²) + ½(0.170)(0.929²) = 0.121170...J

As KEf ≠ KEi, it was not an elastic collision.  ANSWER