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

Answer:

490,400 J

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

$k = \frac{1}{2}mv^{2}$

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

Total kinetic energy is given by

$k = \frac{1}{2}mu^{2}+\frac{1}{2}MU^{2}$

$k = \frac{1}{2}\times1000\times20^{2}+\frac{1}{2}\times1200\times22^{2}$

$k = 200000 + 290400$

k = 490,400 J

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