A 2170 kg car moving east at 10.5 m/s collides with a 3230 kg car moving east. The cars stick together and move east as a unit after the collision at a velocity of 6.00 m/s.

a) What is the velocity of the 3230 kg car before the collision?

2170*10.5E+3230V=(3230+2170)6E

solve for V. Notice its direction has to be zero in N, or S direction.

To find the velocity of the 3230 kg car before the collision, we can use the principle of conservation of momentum.

The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. In other words, the total momentum of the two cars before the collision is equal to the total momentum of the two cars after the collision.

Momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v), so we can express the principle of conservation of momentum as:

m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'

Where:
m1 = mass of the first car (2170 kg)
v1 = velocity of the first car before the collision (unknown)
m2 = mass of the second car (3230 kg)
v2 = velocity of the second car before the collision (unknown)
v1' = velocity of the combined cars after the collision (6.00 m/s)
v2' = velocity of the combined cars after the collision (6.00 m/s)

We can rearrange the equation and solve for v2:

m1 * v1 + m2 * v2 = (m1 + m2) * v1'
v2 = [(m1 + m2) * v1' − m1 * v1] / m2

Plugging in the given values:

v2 = [(2170 kg + 3230 kg) * 6.00 m/s − 2170 kg * 10.5 m/s] / 3230 kg

Simplifying the equation:

v2 = [5400 kg * 6.00 m/s − 22835 kg*m/s] / 3230 kg

v2 = (32400 kg*m/s - 22835 kg*m/s) / 3230 kg

v2 = 9571.47 kg*m/s / 3230 kg

v2 ≈ 2.963 m/s

Therefore, the velocity of the 3230 kg car before the collision was approximately 2.963 m/s.