Question

A car of mass 1200 kg, moving with a speed of 72 mph on a highway,

Answer 1

momentum(p) = mass(m) x velocity(v)

KE = kinetic energy = 1/2 mv²

a.

the ratio of the momentum of the SUV to that of the car

$\tt \dfrac{mv}{1.5m\dfrac{2}{3}v }=1:1$

b.  the ratio of the KE of the SUV to that of the car

$\tt \dfrac{1/2\times mv^2}{1/2\times 1.5m\times (\dfrac{2}{3}v )^2}=\dfrac{mv^2}{1.5m\dfrac{4}{6}v^2 }=1:1$