S4.6 Collisions
[A] Context
- A collision occurs whenever two bodies interact strongly for a short time.
- Collisions can be completely elastic, completely inelastic, or somewhere in between.
- Conservation of linear momentum and where appropriate energy can be very powerful techniques for analysing collisions.
Examples
Reveal

Examples
Collisions range widely in scale, from the impact of comets on planets, through the contact of a tennis ball with a racquet, to the collision of photons with our eyes.
Commentary
Note that a collision does not require a “crash” or physical contact. Eg. When a spacecraft swings round a planet to pickup speed, this can also be regarded as a collision.
Much of modern experimental particle and condensed matter physics relies on collisions or ‘scattering’ between a beam of particles and the physical system of interest. In fact most of our knowledge of the sub-molecular world comes from such experiments. Conservation of linear momentum and energy are key factors in interpreting the results.
[B] Impulse
- In a collision, forces can vary rapidly as a function of time. We use
impulse to quantify the strength and duration of a collision.
Key Point 4.11
Impulse is the change in momentum a particle experiences during a collision
By definition, the impulse is equal to the average force (Key Point 4.7) acting during the collision multiplied by the duration of the collision:
For a 1D collision, the impulse is the area under the F(t) curve describing the collision |
[C] Inelastic collisions
- Collisions in which energy is dissipated are termed inelastic
- The collision of two cars, with the subsequent tearing and deforming of metal, plastic and organic matter dissipates a lot of energy, in work done on the metal, in noise and in heat: car crashes are definitely inelastic.
[D] Elastic collisions
- Collisions in which no energy is dissipated are termed elastic.
- Collisions are elastic, if, for a closed system of two particles the total kinetic energy (in addition to the momentum) is conserved. Using the same notation as above, we have an additional relation expressing conservation of kinetic energy.
- the collision of a hard rubber ball with the floor is almost completely elastic. The ball rises almost to the height from which it was released.
[E] Problem solving tactics for collisions
- Decide on the boundaries of your system.
- Decide if your system is closed and isolated. Closed means no matter passes through the boundaries. Isolated means interactions only occur between particles in your system: the net external force is zero.
- Decide if energy is conserved. In most problems you will meet it will not be.
Remember that in a closed, isolated system, linear momentum is conserved regardless of whether the collision is elastic or inelastic. If particles 1 and 2 have momenta
and
respectively, momentum conservation implies that
where additional subscripts i and f denote initial and final momenta.
- Remember also that linear momentum will be conserved for each of its components. (If the system is not isolated it may be the case that momentum will be conserved in one component only, say the component perpendicular to an external force.)
- Select two states before and after a collision, and equate momenta (and where appropriate energies) before and after the collision. Solve for what is required.

Worked example
Two blocks of mass 300 g and 200 g are moving towards one another along a horizontal frictionless surface with velocities of 50 cm s-1 and 100 cm s-1 respectively.
- [(a)] If the blocks collide and stick together, find their final velocity.
- [(b)] Find the loss of kinetic energy during the collision.
- [(c)] Find the final velocity of each block if the collision is completely elastic.
Solution
(a) Since the surface is frictionless, it is isolated in the direction of motion and so momentum is conserved:
![\[ \vec{p}_{1i}+\vec{p}_{2i}=\vec{p}_{1f}+\vec{p}_{2f} \]](mastermathpng-6.png)
![\begin{eqnarray*} %%%Was makeeqnarraystar; label was blocks1 {\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(0.3 \times 0.5) - (0.2 \times 1.0)}} & = & {\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(0.3+0.2) \times v_f}} \end{eqnarray*}](mastermathpng-7.png)
So vf=-0.1ms-1
(b)
![\begin{eqnarray*} %%%Was makeeqnarraystar; label was blocks2 K_i &=& {\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{0.5\left[0.3\times (0.5)^2 + 0.2\times (1.0)^2\right]}} \\ &=& 0.1375 J \\ K_f &=& {\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{0.5\times (0.3+0.2) \times (-0.1)^2}} \\ &=& 0.0025 J \end{eqnarray*}](mastermathpng-8.png)
Hence the loss in kinetic energy is 0.135J.
(c) For an elastic collision, both momentum and kinetic energy are conserved.

Given m1,m2,v1i,v2i, can solve to find v1f=-0.7ms-1, v2f=0.8ms-1.
Learning Resources
![]() | HRW 9.6 and 9.8-9.11 |
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