Two identical2,000-kg cars travel each at 50.0 km/h in directions perpendicular to one another. The cars crash and form asingle 4,000-kg mass. Comparing the kinetic energy of the two-car system just before the crash and just after it, how much energy was lost due to the collision?

0Joules

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion. It is expressed mathematically as;

K.E = 1/2mv² where

m is the mass of the object

v is the velocity

Given mass of the cars = 2000kg

Velocity = 50km/hr = 50000m/3600s

Velocity = 13.89m/s

Kinetic energy of the 2000kg identical cars traveling at a speed of 13.89m/s before collision is given as;

K.E = 1/2 × 2000 × 13.89²

K.E = 192,932.1Joules

Their individual kinetic energy before collision is 192,932.1Joules

Their total kinetic energy before collision will be 192,932.1+192,932.1

= 385,864.2Joules

To get the kinetic energy of the bodies after collision, we must first know their common velocity after collision.

According to the conservation law which states that 'the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision.

Momentum = mass × velocity

Before collision, momentum of each bodies will be;

2000 × 13.89

= 27,780kgm/s

After collision their momentum will be;

(2000+2000)v

= 4000v kgm/s²

Using the law to calculate v;

27780+27780 = 4000v

55,560 = 4000v

v = 55,560/4000

v = 13.89m/s

Their KE after collision will then be;

KE = 1/2(4000)×13.89²

KE(after) = 385,864.2Joules

Energy lost due to collision will be KE(before collision) - KE(after collision)

Energy lost due to collision = 385,864.2-385,864.2

Energy lost after collision is 0Joules which shows that no energy was lost after collision.

i think the answer is a, hope this is right, that would be my best guess.

3.2x10^7