TASC TCE 2023-style practice: A bullet of mass 15\ g travelling at 350\ m s^-1 hits a metal plate at 30^ to the surface. It bounces off at 30^ to the surface at 200\ m s^-1. Calculate the magnitude of the change in momentum of the bullet.

The change in momentum is the vector difference, p = p_f - p_i, so use components. Take the surface as horizontal (x) and the outward normal as the y axis. Initial momentum (into the plate, 30^ below the surface): p_ix = (0.015)(350)cos 30^ = 4.55\ kg m s^-1; p_iy = -(0.015)(350)sin 30^ = -2.625\ kg m s^-1. Final momentum (away from the plate, 30^ above the surface, same horizontal direction): p_fx = (0.015)(200)cos 30^ = 2.60\ kg m s^-1; p_fy = +(0.015)(200)sin 30^ = +1.50\ kg m s^-1. Change: p_x = 2.60 - 4.55 = -1.95; p_y = 1.50 - (-2.625) = 4.125\ kg m s^-1. $| p| = sqrt((1.95)^2 + (4.125)^2) = sqrt(20.8) = 4.56\ kg m s^-1.$ Markers want both components resolved before combining.

§-Syllabus dot point

TASPhysicsSyllabus dot point

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.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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:

p = mv

It is a vector pointing in the direction of the velocity, measured in $\text{kg m s}^{-1}$ . 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:

F\,\Delta t = \Delta p = m v - m u

This is just Newton's second law rewritten, since $F = ma = m\dfrac{\Delta v}{\Delta t}$ . On a force-time graph the impulse is the area under the curve, which lets you handle forces that vary during the contact.

The impulse-momentum relationship is the bridge between the force-based and energy-based views of mechanics. It explains why a follow-through in sport increases the speed imparted to a ball (a larger contact time means a larger impulse), and why catching a fast object by drawing your hands back reduces the force you feel (a longer time for the same change in momentum).

Conservation of momentum

In a closed system, where no external net force acts, total momentum is constant. For two objects interacting:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

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.

Elastic and inelastic collisions

Momentum is conserved in every collision, but kinetic energy is not. In an elastic collision the total kinetic energy is conserved, as for the billiard balls above. In an inelastic collision some kinetic energy is converted to heat, sound and deformation; in a perfectly inelastic collision the objects stick together and move with one common velocity, losing the maximum possible kinetic energy consistent with conserving momentum. Always test for elasticity by comparing the total kinetic energy before and after, never by assuming.

Two-dimensional momentum

Because momentum is a vector, it is conserved separately in the $x$ and $y$ 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, as in the tangled-players question above.

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.

TCE 20247 marksPlayer A (mass

65\ \text{kg}

) runs at

5.00\ \text{m s}^{-1}

towards north. Player A collides with player B (mass

72\ \text{kg}

) running at

5.50\ \text{m s}^{-1}

towards west, and the two tangle together. Calculate the total momentum of the two players and the final velocity of the entangled players.

Show worked answer →

Momentum is conserved, so the total after the collision equals the vector sum of the two players' momenta before it.

North momentum (A): $p_A = (65)(5.00) = 325\ \text{kg m s}^{-1}$ north.
West momentum (B): $p_B = (72)(5.50) = 396\ \text{kg m s}^{-1}$ west.

These are perpendicular, so add them as a right-angled triangle:

p = \sqrt{(325)^2 + (396)^2} = \sqrt{262441} = 512\ \text{kg m s}^{-1}.

Direction:

\theta = \tan^{-1}\!\left(\dfrac{396}{325}\right) = 50.6^\circ

west of north.

Final velocity: combined mass $65 + 72 = 137\ \text{kg}$ , so $v = \dfrac{p}{m} = \dfrac{512}{137} = 3.74\ \text{m s}^{-1}$ at $50.6^\circ$ west of north. Markers reward the perpendicular vector triangle, the conserved total momentum, and dividing by total mass.

TCE 20235 marksA bullet of mass

15\ \text{g}

travelling at

350\ \text{m s}^{-1}

hits a metal plate at

30^\circ

to the surface. It bounces off at

30^\circ

to the surface at

200\ \text{m s}^{-1}

. Calculate the magnitude of the change in momentum of the bullet.

Show worked answer →

The change in momentum is the vector difference, $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$ , so use components. Take the surface as horizontal ( $x$ ) and the outward normal as the $y$ axis.

Initial momentum (into the plate, $30^\circ$ below the surface):
$p_{ix} = (0.015)(350)\cos 30^\circ = 4.55\ \text{kg m s}^{-1}$ ; $p_{iy} = -(0.015)(350)\sin 30^\circ = -2.625\ \text{kg m s}^{-1}$ .

Final momentum (away from the plate, $30^\circ$ above the surface, same horizontal direction):
$p_{fx} = (0.015)(200)\cos 30^\circ = 2.60\ \text{kg m s}^{-1}$ ; $p_{fy} = +(0.015)(200)\sin 30^\circ = +1.50\ \text{kg m s}^{-1}$ .

Change: $\Delta p_x = 2.60 - 4.55 = -1.95$ ; $\Delta p_y = 1.50 - (-2.625) = 4.125\ \text{kg m s}^{-1}$ .

|\Delta \vec{p}| = \sqrt{(1.95)^2 + (4.125)^2} = \sqrt{20.8} = 4.56\ \text{kg m s}^{-1}.

Markers want both components resolved before combining.

TCE 20224 marksTwo identical billiard balls of mass

0.170\ \text{kg}

collide. Ball A moves at

2.00\ \text{m s}^{-1}

and strikes stationary ball B. The balls separate at

90^\circ

with ball A moving at

1.50\ \text{m s}^{-1}

. Calculate the speed of ball B, and determine whether the collision is elastic.

Show worked answer →

Conservation of momentum (equal masses cancel from each component). The initial momentum lies entirely along A's original line.

With the two final velocities at $90^\circ$ , the momentum vector triangle gives $v_A^2 + v_B^2 = v_\text{initial}^2$ (the initial vector is the hypotenuse):

v_B^2 = (2.00)^2 - (1.50)^2 = 1.75, \quad v_B = 1.32\ \text{m s}^{-1}.

Elastic check (compare kinetic energy). Before: $E_k = \tfrac12(0.170)(2.00)^2 = 0.340\ \text{J}$ .
After: $E_k = \tfrac12(0.170)\big[(1.50)^2 + (1.32)^2\big] = \tfrac12(0.170)(4.00) = 0.340\ \text{J}$ .

Kinetic energy is conserved, so the collision is elastic. Markers reward the right-angle momentum triangle and the explicit kinetic energy comparison.