Conservation of momentum: the total before equals the total after
Definition
Principle of the conservation of momentum
In a closed system, the total momentum before an interaction is equal to the total momentum after it, provided no external resultant force acts.
m1u1+m2u2=m1v1+m2v2 m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 m1u1+m2u2=m1v1+m2v2where uuu values are the velocities before the collision and vvv values are the velocities after it.
- A closed system simply means that you have counted every object involved and that nothing outside the system pushes or pulls on it while the interaction happens.
- The principle holds for collisions, where objects run into each other, and for explosions, where objects push apart from rest.
- Momentum is a vector, so the sums only work if you choose a positive direction first and give anything moving the other way a negative velocity.
- In an explosion the total momentum before is zero, because nothing is moving. The two fragments must therefore fly apart with equal and opposite momenta, which is why a rifle recoils and why a balloon shoots forwards as air rushes backwards.
- When two objects stick together after a collision, treat them afterwards as a single object of combined mass moving with one shared velocity.
- Kinetic energy is not always conserved in a collision, even though momentum always is. Objects that crumple, stick or make a noise lose kinetic energy to internal energy and sound, yet their total momentum is unchanged.
Resultant force is the rate at which momentum changes
Definition
Resultant force
The change in momentum per unit time produced on an object.
F=ΔpΔt F = \frac{\Delta p}{\Delta t} F=ΔtΔpwhere Δp\Delta pΔp is the change in momentum in kg m/s\text{kg}\,\text{m/s}kgm/s and Δt\Delta tΔt is the time taken in s\text{s}s.
- This is the same statement as impulse=FΔt\text{impulse} = F\Delta timpulse=FΔt, written the other way round, and it is the more general form of F=maF = maF=ma.
- If the mass is constant, Δp=mΔv\Delta p = m\Delta vΔp=mΔv, so F=mΔvΔt=maF = \dfrac{m\Delta v}{\Delta t} = maF=ΔtmΔv=ma. The momentum version is more powerful because it still works when mass itself changes, as it does for a rocket burning fuel.
- The equation says that a force does not produce motion by itself: it produces a rate of change of momentum. A steady force therefore builds momentum steadily.
- Because the change in momentum is fixed by the collision, increasing Δt\Delta tΔt decreases FFF. This is the quantitative reason behind safety features that extend an impact.
- The force calculated this way is an average force over the time interval, since the real force during a collision rises and falls.
Common Mistake
- Momentum is conserved; kinetic energy often is not. Saying "energy is conserved so the speeds must stay the same" is wrong: the kinetic energy can drop while momentum stays fixed.
- Signs are not optional. Writing 6.0+4.06.0 + 4.06.0+4.0 for two objects approaching each other, instead of 6.0−4.06.0 - 4.06.0−4.0, is the classic slip. Define positive first.
- Do not forget the objects at rest. They contribute zero momentum, but they still contribute mass to the combined object after a collision.
- F=Δp/ΔtF = \Delta p / \Delta tF=Δp/Δt gives an average force, not the peak force during the impact.
Exam technique
- Start every problem by writing "taking motion to the right as positive" and a before and after sketch with a velocity arrow on each object. Most lost marks here are sign errors, not physics errors.
- Set the working out as total momentum before === total momentum after and substitute into that line, so the examiner can follow each term.
- When two objects join, the mass afterwards is the sum of the masses; when one splits, the masses afterwards must add back to the original.
- A negative answer is a result, not a mistake. Say what it means: "the piece moves in the opposite direction at 7.2 m/s7.2\ \text{m/s}7.2 m/s".
Self review
- State the principle of the conservation of momentum, and the condition under which it applies.
- A 3.0 kg3.0\ \text{kg}3.0 kg trolley at 4.0 m/s4.0\ \text{m/s}4.0 m/s hits a stationary 1.0 kg1.0\ \text{kg}1.0 kg trolley and they stick together. Calculate their common velocity.
- Explain why a gun recoils when it is fired.
- Write the equation that defines resultant force in terms of momentum.
- Explain, using that equation, why airbags reduce the force on a driver.
