Most two-dimensional collision problems are solved by resolving every momentum into two perpendicular components and applying conservation of momentum to each direction separately.Method:
1. Choose axes, usually with x along the initial direction of motion.
2. Resolve each momentum into components: px = p cosθ and py = p sinθ, giving each a sign (+ or –).
3. Write two equations: total px before = total px after, and total py before = total py after.
4. Solve for the unknown components, then combine them using Pythagoras (magnitude) and tan (direction).If an object moving along x hits a stationary object, the total y-momentum before is zero. So after the collision the y-components must add up to zero: the two objects move off on opposite sides of the original line.
Worked example: a 1.0 kg puck moving at 6.0 m s-1 along x hits a stationary 2.0 kg puck. The 1.0 kg puck moves off at 4.0 m s-1 at 60° above the x-axis. Find the velocity of the 2.0 kg puck.
x: 1.0 × 6.0 = 1.0 × 4.0 cos 60° + px, so 6.0 = 2.0 + px and px = 4.0 kg m s-1
y: 0 = 1.0 × 4.0 sin 60° + py, so py = –3.5 kg m s-1
p = √(4.02 + 3.52) = 5.3 kg m s-1, so v = 5.3 / 2.0 = 2.6 m s-1
direction: tan-1(3.5 / 4.0) = 41° below the x-axis.
1. Choose axes, usually with x along the initial direction of motion.
2. Resolve each momentum into components: px = p cosθ and py = p sinθ, giving each a sign (+ or –).
3. Write two equations: total px before = total px after, and total py before = total py after.
4. Solve for the unknown components, then combine them using Pythagoras (magnitude) and tan (direction).If an object moving along x hits a stationary object, the total y-momentum before is zero. So after the collision the y-components must add up to zero: the two objects move off on opposite sides of the original line.
Worked example: a 1.0 kg puck moving at 6.0 m s-1 along x hits a stationary 2.0 kg puck. The 1.0 kg puck moves off at 4.0 m s-1 at 60° above the x-axis. Find the velocity of the 2.0 kg puck.x: 1.0 × 6.0 = 1.0 × 4.0 cos 60° + px, so 6.0 = 2.0 + px and px = 4.0 kg m s-1
y: 0 = 1.0 × 4.0 sin 60° + py, so py = –3.5 kg m s-1
p = √(4.02 + 3.52) = 5.3 kg m s-1, so v = 5.3 / 2.0 = 2.6 m s-1
direction: tan-1(3.5 / 4.0) = 41° below the x-axis.