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Momentum and impulse - TCE Physics (Tasmania)

Syllabus dot point

“Apply conservation of momentum and the impulse-momentum relationship to collisions and explosions in one and two dimensions.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

Momentum p=mvp = mv is conserved in any closed system, so the total momentum before a collision or explosion equals the total after. Impulse is the change in momentum, FΔt=ΔpF\Delta t = \Delta p, so a force acting over a time interval transfers momentum.

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=mvp = mv

It is a vector pointing in the direction of the velocity, measured in kg m s−1\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 Δt=Δp=mv−muF\,\Delta t = \Delta p = m v - m u

This is just Newton's second law rewritten, since F=ma=mΔvΔtF = 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.

Impulse

Impulse is the product of the average force on an object and the time for which it acts, equal to the change in the object's momentum. It is a vector measured in newton seconds, equivalent to kilogram metres per second.

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:

m1u1+m2u2=m1v1+m2v2m_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 xx and yy 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.

Worked example: a perfectly inelastic collision

Set up conservation of momentum

A 1200 kg1200\ \text{kg} car moving east at 15 m s−115\ \text{m s}^{-1} collides and sticks to a stationary 800 kg800\ \text{kg} car. Take east as positive:

m1u1+m2u2=(m1+m2)v.m_1 u_1 + m_2 u_2 = (m_1 + m_2)v.

Solve for the common velocity

(1200)(15)+(800)(0)=(1200+800)v,18000=2000v⇒v=9.0 m s−1.(1200)(15) + (800)(0) = (1200 + 800)v, \quad 18000 = 2000v \Rightarrow v = 9.0\ \text{m s}^{-1}.

The joined wreck moves east at 9.0 m s−19.0\ \text{m s}^{-1}.

Impulse explains safety design

For a given change in momentum, extending the collision time reduces the average force, since F=ΔpΔtF = \dfrac{\Delta p}{\Delta t}. Crumple zones, airbags and padded surfaces all increase Δt\Delta t to lower the peak force on the body.

Forgetting that momentum is a vector

A common slip is adding momenta as if they were numbers without regard to direction. Two objects moving toward each other have opposite signs, so their total momentum is the difference, not the sum. Always assign a positive direction and give each momentum the correct sign before adding.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks
Player A (mass 65 kg65\ \text{kg}) runs at 5.00 m s−15.00\ \text{m s}^{-1} towards north. Player A collides with player B (mass 72 kg72\ \text{kg}) running at 5.50 m s−15.50\ \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): pA=(65)(5.00)=325 kg m s−1p_A = (65)(5.00) = 325\ \text{kg m s}^{-1} north.
West momentum (B): pB=(72)(5.50)=396 kg m s−1p_B = (72)(5.50) = 396\ \text{kg m s}^{-1} west.

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

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

Direction: θ=tan⁡−1 ⁣(396325)=50.6∘\theta = \tan^{-1}\!\left(\dfrac{396}{325}\right) = 50.6^\circ west of north.

Final velocity: combined mass 65+72=137 kg65 + 72 = 137\ \text{kg}, so v=pm=512137=3.74 m s−1v = \dfrac{p}{m} = \dfrac{512}{137} = 3.74\ \text{m s}^{-1} at 50.6∘50.6^\circ west of north. Markers reward the perpendicular vector triangle, the conserved total momentum, and dividing by total mass.

Practice5 marks
A bullet of mass 15 g15\ \text{g} travelling at 350 m s−1350\ \text{m s}^{-1} hits a metal plate at 30∘30^\circ to the surface. It bounces off at 30∘30^\circ to the surface at 200 m s−1200\ \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, Δp⃗=p⃗f−p⃗i\Delta \vec{p} = \vec{p}_f - \vec{p}_i, so use components. Take the surface as horizontal (xx) and the outward normal as the yy axis.

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

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

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

∣Δp⃗∣=(1.95)2+(4.125)2=20.8=4.56 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.

Practice4 marks
Two identical billiard balls of mass 0.170 kg0.170\ \text{kg} collide. Ball A moves at 2.00 m s−12.00\ \text{m s}^{-1} and strikes stationary ball B. The balls separate at 90∘90^\circ with ball A moving at 1.50 m s−11.50\ \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∘90^\circ, the momentum vector triangle gives vA2+vB2=vinitial2v_A^2 + v_B^2 = v_\text{initial}^2 (the initial vector is the hypotenuse):

vB2=(2.00)2−(1.50)2=1.75,vB=1.32 m s−1.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: Ek=12(0.170)(2.00)2=0.340 JE_k = \tfrac12(0.170)(2.00)^2 = 0.340\ \text{J}.
After: Ek=12(0.170)[(1.50)2+(1.32)2]=12(0.170)(4.00)=0.340 JE_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.

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Sources & how we know this

Based on the official TASC course document. Not yet individually reviewed by a person; always check TASC and your school's materials for current rules.

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