Question

Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmoves forward at 25 m/s and car A is at rest. What fraction of theinitial kinetic energy is lost in the collision?
My approach:
How exactly do you approach this question? At first, I thoughtusing the the formula ΣKEi = ΣKEf.Then, to find the fraction just divide the former from the latter.But is it that cut and dry?
Before collision:
viA= 35 m/s
viB = 0 m/s
vfA = 0 m/s
vfB =25 m/s
m1 = m2

Answer 1

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

$ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2$

$kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2$

Since the masses are equals:

$m=ma=mb$

For the known velocities, the kinetics energies result:

$ki=0.5*mVa_{i}^2$

$ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m$

$kf=0.5*mbVb_{f}^2$

$kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m$

The lost energy in the collision is the difference between the initial and final kinectic energies:

$kl=ki-kf$

$kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m$

Finally the relation between the lost and the initial kinetic energy:

$kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m$

$kl/ki = 24/49=0.49$