4.3 Momentum
- Syllabus
- 9709–2028–2029
- Topic
- 4.3
- Level
- A2
Learning objectives
4.3.1Momentum• use the definition of linear momentum and show understanding of its vector nature For motion in one dimension only.4.3.2Momentum• use conservation of linear momentum to solve problems that may be modelled as the direct impact of two bodies. Including direct impact of two bodies where the bodies coalesce on impact. Knowledge of impulse and the coefficient of restitution is not required.
Momentum is mass multiplied by signed velocity
Forone−dimensionalmotion,p=mv,where $m>0$ and velocity $v$ carries the chosen-direction sign. Units are kg m s$^{-1}$.
Choose a positive direction once. Momentum is positive or negative according to velocity; reversing motion reverses momentum without changing mass.
A $2$ kg particle moving at $+4$ m s$^{-1}$ has $p=+8$ kg m s$^{-1}$; a $1$ kg particle moving at $-1$ m s$^{-1}$ has $p=-1$ kg m s$^{-1}$.
Momentum is a vector quantity restricted here to one dimension, so it is represented by a signed scalar. Its magnitude is m times speed.
Do not replace velocity by speed when direction matters. Conservation belongs to the next objective, not to the definition of one particle’s momentum.
Balance signed momentum before and after a direct impact
Foramodelleddirectimpactwithnegligibleexternaleffectduringtheevent:m_1u_1+m_2u_2=m_1v_1+m_2v_2.Everyvelocityissignedinonechosendirection.
Define the two-body system and positive direction, label velocities immediately before/after, write one signed momentum equation, include any stated relation between final velocities, solve and interpret a negative result as opposite to the assumed direction.
If the bodies coalesce, $v_1=v_2=v$:m_1u_1+m_2u_2=(m_1+m_2)v.
A $2$ kg body at $4$ m s$^{-1}$ hits a $1$ kg body at $-1$ m s$^{-1}$ and they stick:2(4)+1(-1)=3v\Rightarrow v=\frac73\text{ m s}^{-1}.
Momentum conservation does not imply kinetic-energy conservation. Knowledge of impulse and coefficient of restitution is explicitly not required.