A 2189 kg car moving east at 9.73 m/s collides with a 3285 kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 35.7◦ north of east and at a speed of 4.79 m/s.

What was the speed of the 3285 kg car
before the collision?
Answer in units of m/s

Use the equation m1u1 + m2u2 = m1v1 + m2v2

(mass 1 x its initial velocity + mass 2 x its initial velocity = mass 1 x its final velocity + mass 2 x its final velocity)

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To solve this problem, we can use the principles of conservation of momentum and trigonometry.

Conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision. Therefore, we can set up the equation:

(mass of car 1) * (velocity of car 1) + (mass of car 2) * (velocity of car 2) = (total mass) * (final velocity)

Let's assume that the initial velocity of the 3285 kg car is v1, and the final velocity after the collision is v2.

We can rewrite the equation as:

(2189 kg) * (9.73 m/s) + (3285 kg) * v1 = (2189 kg + 3285 kg) * (4.79 m/s)

Now, we can solve this equation for v1:

(2189 kg) * (9.73 m/s) + (3285 kg) * v1 = (5474 kg) * (4.79 m/s)
21316.97 kg*m/s + 3285 kg * v1 = 26223.46 kg*m/s
3285 kg * v1 = 4910.49 kg*m/s
v1 = 4910.49 kg*m/s / 3285 kg
v1 = 1.494 m/s

Therefore, the speed of the 3285 kg car before the collision was approximately 1.494 m/s.