A 1 145.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east. What is the velocity of the truck right after the collision?(I found this) =20.826 (b) What is the change in mechanical energy of the car?

(a)v2 = 20.590m/s

(b) ΔE = 116,150J

Explanation:

Let m1 = mass of the car = 1145kg

m2 = mass of the truck = 9700kg

u1 = 25.000m/s = initial velocity of the car

u2 = 20.000m/s = initial velocity of the truck

v1 = 20.000m/s = final velocity of the car

v2 = ? = final velocity of the truck

From the principle of the conservation of momentum, total momentum before collision equals total momentum after collision. We sum the momentum of both vehicles before and after the collision. For the car the momentum before and after the collision are m1u1 and m1v1 while for the truck they are m2u2 and m2v2

So,

m1u1 + m2u2 = m1v1 + m2v2

1145×25 + 9700×20 = 1145×20 + 9700v2

28,625 + 194,000 = 22,900 + 9700v2

222,625 = 22,900 +9700v2

9700v2 = 222625 –22900

9700v2 = 199,725

v2= 199725/9700 = 20.59m/s

(b) ΔE = 1/2× m2× (v2²– v1²)

= 1/2×9700×(20.590² – 20.000²)

= 116,148J

reserve is the answer