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.

Question

What is the speed and direction of the wreckage immediately after impact?

Answer 1

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.