Final answer:
When the 250 kg car traveling at 1.5 m/s collides head-on with the 150 kg car traveling at 2.5 m/s, and both cars come to rest with the springs compressed by 17 cm, the average force exerted on each car during the collision is approximately 13,235 N.
Step-by-step explanation:
To calculate the average force exerted on each car during the collision, we can apply the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision.
The momentum (p) of each car before the collision is given by the product of its mass (m) and velocity (v). The total initial momentum of the system is the sum of the individual momenta of the two cars before the collision.
After the collision, both cars come to rest, and the kinetic energy of the cars is transferred into the potential energy stored in the compressed springs. The potential energy stored in each spring can be determined using the formula for spring potential energy:
where k is the spring constant and x is the compression of the spring.
Given that the springs are compressed by 17 cm (0.17 m) when the cars come to rest, and the springs are identical, we can find the spring constant using Hooke's Law:
where F is the force applied, and x is the compression. Once the spring constant is determined, we can calculate the potential energy stored in the springs.
Finally, using the principle of conservation of energy, the kinetic energy before the collision equals the total potential energy stored in the compressed springs after the collision. By equating these energies and solving for the force applied during compression, we find that the average force exerted on each car during the collision is approximately 13,235 N.