Car A, with a mass of 1250 kg, is traveling at 30 m/s to the east. Car B is a truck with a mass of 2000 kg, traveling to the west at 25 m/s. Assume these two vehicles experience an inelastic collision but do not stick together, and Car A goes off 10 m/s to the west. What will be the resulting velocity of Car B?

Since we're dealing with just east/west, we can use just +/- for our velocities, with + meaning east.

pA = 1250(30) = 37500
pB = 2000(-25) = -50000
sum = -12500

The momentum of the system remains unchanged. After the collision,

pA = 1250(-10) = -12500
Since A has all the momentum now, B has come to a stop.

Hi, I was wondering if anyone could answer my question. A two head collision. One car is 45000 newtons and going at 15 meters per second. The other car is 9000 newtons and going at 25 meters per second. How would I calculate velocity from this.

To determine the resulting velocity of Car B after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.

The momentum of an object is calculated by multiplying its mass by its velocity. Let's denote the velocity of Car B after the collision as VBf.

The initial momentum of Car A (before the collision) is given by the product of its mass (mA) and its initial velocity (VAi):

Momentum of Car A before collision = mA × VAi

The initial momentum of Car B (before the collision) is given by the product of its mass (mB) and its initial velocity (VBi):

Momentum of Car B before collision = mB × VBi

After the collision, Car A goes off with a velocity of -10 m/s (to the west). The negative sign indicates that its direction has changed. Since the two vehicles experience an inelastic collision but do not stick together, Car B must have a positive velocity afterward (to the west).

Considering conservation of momentum, we have:

Momentum of Car A before collision + Momentum of Car B before collision = Momentum of Car A after collision + Momentum of Car B after collision

(mA × VAi) + (mB × VBi) = (mA × VAf) + (mB × VBf)

Plugging in the given values:

(1250 kg × 30 m/s) + (2000 kg × (-25 m/s)) = (1250 kg × (-10 m/s)) + (2000 kg × VBf)

Simplifying the equation:

37500 kg·m/s - 50000 kg·m/s = -12500 kg·m/s + 2000 kg × VBf

-12500 kg·m/s - VBf = -50000 kg·m/s - 37500 kg·m/s

-12500 kg·m/s - VBf = -87500 kg·m/s

VBf = -87500 kg·m/s + 12500 kg·m/s

VBf ≈ -75000 kg·m/s

Therefore, the resulting velocity of Car B after the collision is approximately -75000 m/s (to the west).