Abumper car with a mass of 200 kilograms traveling at a velocity of 3.5 meters/second collides with a stationary bumper car with a mass of 250 kilograms. if the two bumper cars stick and move together at a velocity of 2.3 meters/second, what is the kinetic energy of the system after the collision? 4.5 × 10^2 joules 6.0 × 10^2 joules 8.7 × 10^2 joules 1.2 × 10^3 joules

fourth option 1.2 × 10³ joules

Explanation:

1) kinetic energy is calculated as the half of the product of the mass times the square of the speed:

=> KE = (1/2) m × (v)²

2) you know the mass after the collision and the speed so you have all the data fo find the numerical value of the expression for KE:

m = m1 + m2 = 200 kg + 250 kg = 450 kg

KE = (1/2) × 450 kg × (2.3 m/s)²  = 1,190.25 joules ≈ 1200 joules = 1.2 × 10³ joules, which is the last option of the list.

B) as the temperature increases, the kinetic energy of the particles increases.|-/