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 principles of conservation of momentum and breakdown the components of the velocities.
Let's denote the speed of the 3221 kg car before the collision as "v".
Before the collision, the momentum of the system is conserved. In other words, the total momentum of the cars before the collision is equal to the total momentum of the cars after the collision.
The momentum of an object is given by the formula: momentum = mass × velocity.
The momentum of the 2191 kg car moving east is given by: momentum1 = (mass1) × (velocity1) = (2191 kg) × (9.9 m/s)
Now, let's break down the velocity of the 3221 kg car moving north into its components.
Given that the final velocity of the combined cars makes an angle of 37.5° north of east, we can determine the northward component of the velocity using trigonometry.
northward component = velocity × sin(angle) = v × sin(37.5°).
Since the cars stick together and move as a unit, the mass of the combined system is (2191 kg + 3221 kg).
The momentum of the combined system after the collision is given by: momentum2 = (mass1 + mass2) × (velocity2) = (2191 kg + 3221 kg) × (5.05 m/s)
According to the conservation of momentum, momentum1 = momentum2.
Therefore, we can equate the two equations:
(mass1) × (velocity1) = (mass1 + mass2) × (velocity2)
(2191 kg) × (9.9 m/s) = (2191 kg + 3221 kg) × (5.05 m/s)
Now, we can solve this equation to find the value of "v", the initial speed of the 3221 kg car:
(2191 kg) × (9.9 m/s) = (2191 kg + 3221 kg) × (5.05 m/s)
Solving for "v" will give us the answer to the question.