Caden is driving a corvette with a mass of 2.5 x 10^3 kg traveling west at 6.0 m/s is hit by Mason's truck with a mass of 6.0 x 10^3 kg going south at 4.0 m/s. The two vehicles lock together on impact.
What is the speed and direction of the wreckage immediately after impact?
To solve this problem, we can use the principle of conservation of momentum.
First, calculate the total momentum of the system before the collision:
Total momentum before collision = (mass of Corvette x velocity of Corvette) + (mass of truck x velocity of truck)
Total momentum before collision = (2.5 x 10^3 kg) x (6.0 m/s) + (6.0 x 10^3 kg) x (4.0 m/s)
Total momentum before collision = 15,000 kg m/s + 24,000 kg m/s
Total momentum before collision = 39,000 kg m/s
Now, since the two vehicles lock together after the collision, the total momentum after the collision will be the same as before the collision. Let the speed of the wreckage be v and the direction be θ.
Total momentum after collision = (total mass of wreckage x speed of wreckage)
Total momentum after collision = (2.5 x 10^3 kg + 6.0 x 10^3 kg) x v
Total momentum after collision = 8.5 x 10^3 kg x v
Now, using the principle of conservation of momentum:
Total momentum before collision = Total momentum after collision
39,000 kg m/s = 8.5 x 10^3 kg x v
v = 4.59 m/s
Now, to find the direction, we can use vector addition. The final velocity of the wreckage will be in the direction that makes the total momentum equal to the initial momentum.
tanθ = (24,000 kg m/s) / (15,000 kg m/s)
θ = atan(24,000 / 15,000)
θ = atan(1.6)
θ = 57.99 degrees
Therefore, the speed of the wreckage immediately after impact is 4.59 m/s and the direction is approximately 58 degrees south of west.