Question

A 1000-kg car is traveling east at 20\:m.s^{-1}20 m . s − 1 and a 1200-kg car is traveling west at 22\:m.s^{-1}\:22 m . s − 1. What is the total kinetic energy of the two-car system in the center of mass reference frame? (i)\:\:\:\:\:3.92\:\times\:10^5\:J

Answer 1

490,400 J

Step-by-step explanation:

Mass of first car, m = 1000 kg

Mass of second car, M = 1200 kg

velocity of first car, u = 20 m/s east

velocity of second car, U = 22 m/s west

The formula for the kinetic energy is

`k = (1)/(2)mv^(2)`

where, m is the body and v be the velocity of the body.

Total kinetic energy is given by

`k = (1)/(2)mu^(2)+(1)/(2)MU^(2)`

`k = (1)/(2)*1000*20^(2)+(1)/(2)*1200*22^(2)`

`k = 200000 + 290400`

k = 490,400 J

Thus, the total kinetic energy of the system is 490,400 J.