a 2191 kg car moving east at 9.9 m/s collides with a 3221 kg car moving north. the cars stick together and move as a unit after the collision, at an angle of 37.5 north of east at a speed of 5.05 m/s. what is the speed of the 3221 kg car before the collision?

To find the speed of the 3221 kg car before the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Given:
Mass of the first car (m1) = 2191 kg
Initial velocity of the first car (v1) = 9.9 m/s (moving east)
Mass of the second car (m2) = 3221 kg
Initial velocity of the second car (v2) = unknown
Final velocity (v') = 5.05 m/s (with an angle of 37.5° north of east)

The momentum of an object is given by the product of its mass and velocity. Therefore, the total momentum before the collision is:

Total momentum before = m1 * v1 + m2 * v2

After the collision, the two cars stick together and move as a unit. The final momentum of the colliding cars is:

Total momentum after = (m1 + m2) * v'

Since the total momentum before the collision is equal to the total momentum after the collision, we can equate the two expressions:

m1 * v1 + m2 * v2 = (m1 + m2) * v'

Now we can plug in the given values:

(2191 kg * 9.9 m/s) + (3221 kg * v2) = (2191 kg + 3221 kg) * 5.05 m/s

(21690.9 kg·m/s) + (3221 kg * v2) = (5412 kg) * 5.05 m/s

21690.9 kg·m/s + 3221 kg * v2 = 27390.6 kg·m/s

To find the value of v2, we need to rearrange the equation:

3221 kg * v2 = 27390.6 kg·m/s - 21690.9 kg·m/s

3221 kg * v2 = 5700.7 kg·m/s

Divide both sides of the equation by 3221 kg:

v2 = 5700.7 kg·m/s / 3221 kg

v2 ≈ 1.77068 m/s

Therefore, the speed of the 3221 kg car before the collision is approximately 1.77068 m/s.