Example 1
A 2.00kg trolley moving east at 3.00ms−1 sticks to a 1.00kg trolley moving north at 4.00ms−1. Neglect external impulse.
px=6.00 and py=4.00kgms−1. Total mass 3.00kg gives vx=2.00 and vy=1.33ms−1. Speed =2.40ms−1,33.7∘ north of east.
BUILD FLUENCY. EARN YOUR ASCENT.Sign in
Year 12 · SACE Stage 2
Momentum conservation in two dimensions
Apply independent component balances to collisions and explosions.
Method and worked examples
For a system with negligible external impulse, total vector momentum is conserved. Internal action–reaction forces cancel in the system balance. Choose common x and y axes, resolve all initial and final momenta, and conserve each component separately. Recombine components with Pythagoras and trigonometry; do not add magnitudes for different directions.
Objects that stick together share a final velocity, so divide the final total momentum components by their combined mass. Momentum conservation does not require kinetic-energy conservation: an inelastic collision can transfer kinetic energy to deformation and internal energy. In an explosion, stored energy can increase kinetic energy while the vector momentum total stays unchanged.
Example 1
Example 2
A stationary object separates into fragments with momenta (3,4) and an unknown vector, in kgms−1.
Initial total momentum is zero, so the second momentum is (−3,−4)kgms−1. Its magnitude is 5kgms−1.
Watch out: Momentum is conserved in an isolated system; kinetic energy is conserved only in elastic interactions, though total energy is always conserved.