What stays the same when objects collide?
Apply conservation of momentum and the impulse-momentum relationship to collisions and explosions in one and two dimensions.
Linear momentum, the impulse-momentum theorem, and conservation of momentum applied to collisions and explosions in one and two dimensions.
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What this dot point is asking
This dot point is about momentum, the quantity that is conserved whenever objects interact through internal forces. It explains collisions, explosions, recoil and why crumple zones save lives.
Momentum and impulse
The momentum of an object is the product of its mass and velocity:
It is a vector pointing in the direction of the velocity, measured in . To change an object's momentum you must apply a force over a time interval. This product is the impulse, and it equals the change in momentum:
This is just Newton's second law rewritten, since . On a force-time graph the impulse is the area under the curve.
Conservation of momentum
In a closed system, where no external net force acts, total momentum is constant. For two objects interacting:
This works because the forces the objects exert on each other are a Newton's third law pair, equal and opposite, so they change each object's momentum by equal and opposite amounts that cancel in the total.
Explosions and recoil are the same idea run in reverse: a system starts at rest with zero total momentum, so the fragments must fly apart with momenta that sum to zero. A rifle recoils because the bullet carries forward momentum that the rifle must balance.
Two dimensional momentum
Because momentum is a vector, it is conserved separately in the and directions. For a two dimensional collision, resolve every velocity into components, then write a conservation equation for each axis and solve the two equations together. This is how you handle objects that collide and move off at angles.
In the exam, define your positive direction, write the total momentum before and after, and set them equal. For collisions where objects stick together, the combined mass moves with one velocity; for explosions, the total before is usually zero. Resolve into components only when motion is genuinely two dimensional.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 TASC7 marksFootball player A (Alice) has mass 65 kg and is running at 5.00 m s-1 towards north. Player A meets player B (Beatrix), mass 72 kg, running at 5.50 m s-1 towards west. After colliding the two players tangle together. Calculate the total momentum of the two players (with a vector diagram) and the final velocity of the entangled players.Show worked answer →
Momentum is conserved, so the total momentum after the collision equals the vector sum of the two players' momenta before it.
North momentum (A): p_A = 65 x 5.00 = 325 kg m s-1 (north).
West momentum (B): p_B = 72 x 5.50 = 396 kg m s-1 (west).
These are perpendicular, so add them as a right-angled vector triangle:
magnitude p = sqrt(325^2 + 396^2) = sqrt(105625 + 156816) = sqrt(262441) = 512 kg m s-1.
Direction: theta = arctan(396 / 325) = 50.6 degrees west of north.
Final velocity: the players move off together with combined mass 65 + 72 = 137 kg, so
v = p / m = 512 / 137 = 3.74 m s-1 at 50.6 degrees west of north.
Markers reward the perpendicular vector triangle, the conserved total momentum, and dividing by total mass for the common final velocity.
2023 TASC5 marksA bullet of mass 15 g travelling at 350 m s-1 hits a metal plate at 30 degrees to the surface. It bounces off at 30 degrees to the surface at a speed of 200 m s-1. Calculate the magnitude of the change in momentum of the bullet.Show worked answer →
Change in momentum is the vector difference, delta p = p_final - p_initial, so it must be found with components. Take the plate surface as horizontal and the outward normal as the y axis.
Initial momentum (into the plate, 30 degrees below the surface):
p_ix = 0.015 x 350 x cos30 = 4.55 kg m s-1 (along surface)
p_iy = -0.015 x 350 x sin30 = -2.625 kg m s-1 (into plate)
Final momentum (away from the plate, 30 degrees above the surface, same horizontal direction):
p_fx = 0.015 x 200 x cos30 = 2.60 kg m s-1
p_fy = +0.015 x 200 x sin30 = +1.50 kg m s-1
Change: delta p_x = 2.60 - 4.55 = -1.95 kg m s-1; delta p_y = 1.50 - (-2.625) = 4.125 kg m s-1.
Magnitude = sqrt(1.95^2 + 4.125^2) = sqrt(3.80 + 17.0) = sqrt(20.8) = 4.56 kg m s-1.
The component along the surface changes because the bullet slows; the normal component reverses. Markers want both components resolved before combining.
2022 TASC4 marksTwo identical billiard balls of mass 0.170 kg collide. Ball A is moving at 2.00 m s-1 and strikes a stationary ball B. The balls separate at 90 degrees with ball A moving at 1.50 m s-1. Calculate the speed of ball B, and determine whether the collision is elastic.Show worked answer →
Conservation of momentum (the equal masses cancel from each component). The initial momentum lies entirely along A's original line.
With the two final velocities at 90 degrees to each other, the momentum vector triangle gives:
v_A^2 + v_B^2 = v_initial^2 (the initial vector is the hypotenuse).
So v_B^2 = 2.00^2 - 1.50^2 = 4.00 - 2.25 = 1.75, giving v_B = 1.32 m s-1.
Elastic check: compare kinetic energy. Before: KE = 0.5 x 0.170 x 2.00^2 = 0.340 J.
After: KE = 0.5 x 0.170 x (1.50^2 + 1.32^2) = 0.5 x 0.170 x (2.25 + 1.75) = 0.5 x 0.170 x 4.00 = 0.340 J.
Kinetic energy is conserved, so the collision is elastic. Markers reward the right-angle momentum triangle and the explicit KE comparison.