Conservation of momentum

In all isolated systems, total momentum is conserved:

ptot=ptot\displaystyle \vec{p}_{\text{tot}} = \vec{p}^{\,\prime}_{\text{tot}}.


Car A (2.5 kg) begins at 1.5 m/s [fwd], while Car B (1.5 kg) beings at rest. Car A collides into Car B, and they stick together. What is the velocity of AB after the collision?

We begin with the conservation of momentum,

ptot=ptot\displaystyle \vec{p}_{\text{tot}} = \vec{p}^{\,\prime}_{\text{tot}},

into which we substitute the individual momenta:

pA+pB=pAB\displaystyle \vec{p}_{\text{A}} + \vec{p}_{\text{B}} = \vec{p}^{\,\prime}_{\text{AB}}.

Now we can substitute the masses and velocities with

mAvA+mBvB=mABvAB\displaystyle m_{\text{A}} \vec{v}_{\text{A}} + m_{\text{B}} \vec{v}_{\text{B}} = m_{\text{AB}} \vec{v}^{\,\prime}_{\text{AB}},

and solve for the final velocity:

vAB=mAvA+mBvBmAB\displaystyle \vec{v}^{\,\prime}_{\text{AB}} = \frac{m_{\text{A}} \vec{v}_{\text{A}} + m_{\text{B}} \vec{v}_{\text{B}}}{m_{\text{AB}}}.

Substituting the known values where [fwd] is positive, we have

vAB=(2.5kg)(1.5m/s)02.5kg+1.5kg=0.9375m/s\displaystyle \vec{v}^{\,\prime}_{\text{AB}} = \frac{\left ( 2.5 \, \text{kg} \right ) \left ( 1.5 \, \text{m/s} \right ) - 0}{2.5 \, \text{kg} + 1.5 \, \text{kg}} = 0.9375 \, \text{m/s},

therefore the final velocity is 0.94 m/s [fwd].