A 306-kg car moving at 16.5 m/s hits from behind a 810-kg car moving at 13.2 m/s in the same direction. If the new speed of the heavier car is 17.5 m/s, what is the velocity of the lighter car after the collision, assuming that any unbalanced forces on the system are negligibly small?

Assuming no external forces acting during the collision, total momentum must be conserved.
⇒ p₀ = pf
The initial momentum p₀, can be written as follows:

[tex]p_{o} = m_{1} *v_{1o} + m_{2} *v_{2o} (1)[/tex]

where m₁ = 306 kg, m₂ = 810 kg, v₁₀ = 16.5 m/s, v₂₀ = 13.2 m/s

The final momentum, pf, can be written as follows:

[tex]p_{f} = m_{1} *v_{1f} + m_{2} *v_{2f} (2)[/tex]

where v₂f = 17.5 m/s

Since p₀ = pf, which means that (1) is equal to (2),
Replacing by the givens, and rearranging, we can solve for the only unknown that still remains, v₁f, as follows:

[tex]v_{1f} = v_{1o} +\frac{m_{2} }{m_{1}} * (v_{2o} - v_{2f} ) \\= 16.5 m/s + \frac{810}{306} * (13.2 m/s - 17.5 m/s) \\= 16.5 m/s + \frac{810}{306} * (-4.3 m/s) \\= 16.5 m/s -11.4 m/s = 5.1 m/s[/tex]

The velocity of the lighter car after the collision is 5.1 m/s.