Iwil a 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. after the collision, the 2.00-kg object has a velocity 5.00 m/s to the west. how much kinetic energy was lost during the collision? .

Using the following given values:

Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s 

Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s 

Obtaining Va2 via law of conservation of momentum:

total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67

Total kinetic energy before collision:

KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J

Total kinetic energy after collision:

KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J

Total kinetic energy lost:

Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J

Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy