Conservation of momentum in two dimensions

Momentum is a vector

Total momentum is found by vector addition, not by adding numbers. If no external resultant force acts, the total momentum vector is the same before and after.

Resolve, conserve, recombine

• Choose two perpendicular directions (usually x along the original motion and y at 90° to it).
• Resolve every momentum: p cos θ and p sin θ.
• Conserve momentum separately in x and in y.
• Combine the unknown components with Pythagoras and tan^-1 to give a magnitude and a direction.

Explosions

If an object at rest breaks into pieces, the total momentum is zero in every direction. For three pieces, the third momentum is equal and opposite to the vector sum of the other two.

Worked example

A 2.0 kg puck moving east at 3.0 m/s hits a stationary 1.0 kg puck. Afterwards the 2.0 kg puck moves at 2.0 m/s at 30° north of east. Find the velocity of the 1.0 kg puck.

1. x: 2.0 x 3.0 = 2.0 x 2.0 cos 30° + 1.0 v_x, so v_x = 6.0 - 3.46 = 2.54 m/s
2. y: 0 = 2.0 x 2.0 sin 30° + 1.0 v_y, so v_y = -2.0 m/s
3. v = sqrt(2.54² + 2.0²) = 3.23 m/s; angle = tan^-1(2.0 / 2.54) = 38.3°

Answer: 3.2 m/s at 38° south of east

Check yourself

1. A 3.0 kg object moves at 4.0 m/s at 60° to the x-axis. Find the x-component of its momentum in kg m/s.
   Show answer

   6 kg m/s. p_x = mv cos θ = 12 x cos 60° = 6.0 kg m/s.

2. A stationary object explodes into three pieces. One has momentum 6.0 kg m/s east and another 8.0 kg m/s north. Find the size of the third piece's momentum in kg m/s.
   Show answer

   10 kg m/s. It must cancel the other two: sqrt(6.0² + 8.0²) = 10 kg m/s.

3. In that explosion, the third piece has a mass of 2.5 kg. Find its speed in m/s.
   Show answer

   4 m/s. v = p / m = 10 / 2.5 = 4.0 m/s.

4. In that explosion, find the angle between the third piece's direction and due west, in degrees, to 3 s.f.
   Show answer

   53.1 °. It moves south of west with tan θ = 8.0 / 6.0, so θ = 53.1°.

5. A 0.20 kg ball moving at 5.0 m/s along x hits an identical stationary ball. The first ball moves off at 3.0 m/s at 50° to x. Find the speed of the second ball in m/s, to 3 s.f.
   Show answer

   3.84 m/s. x: 1.0 = 0.386 + p_x, so p_x = 0.614 kg m/s. y: 0 = 0.460 + p_y, so p_y = -0.460 kg m/s. p = sqrt(0.614² + 0.460²) = 0.767 kg m/s; v = 0.767 / 0.20 = 3.84 m/s.

6. How is momentum conserved in a two-dimensional collision?
   • Only in the direction of the first object's motion
   • Separately in each of two perpendicular directions
   • Only as a total magnitude, ignoring direction
   • Only if the collision is elastic
   Show answer

   Separately in each of two perpendicular directions. Momentum is a vector; each perpendicular component is conserved on its own.

7. Two pieces fly apart with momenta 3 kg m/s north and 4 kg m/s east. What is the size of their total momentum?
   • 7 kg m/s
   • 1 kg m/s
   • 12 kg m/s
   • 5 kg m/s
   Show answer

   5 kg m/s. Add as vectors: sqrt(3² + 4²) = 5 kg m/s.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)

Last updated 7 November 2026. Spotted a mistake? Tell us.