A 853-kg car stopped at an intersection is rear-ended by a 1450-kg truck moving with a speed of 16.0 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.

For an elastic collision, both momentum and kinetic energy are conserved.

Conservation of momentum:

m1*v1 + m2*v2 = m1*v1f + m2*v2f

where m1 and m2 are the mass of the car and truck, v1 and v2 are the initial speeds of mass 1 and mass 2, and v1f and v2f are the final speed of mass one and mass 2

Conservation of kinetic energy:
1/2*m1*v1^2 + 1/2*m2*v2^2 =1/2*m1*v1f^2 + 1/2*m2*v2f^2

Insert values, and use algebra to solve for v1f and v2f

To find the final speeds of both vehicles after the collision, we can use the principle of conservation of linear momentum.

The linear momentum before the collision is equal to the linear momentum after the collision. Mathematically, this can be expressed as:

(m1 * v1) + (m2 * v2) = (m1 * v1f) + (m2 * v2f)

where:
m1 and m2 are the masses of the car and truck, respectively.
v1 and v2 are the initial velocities of the car and truck, respectively.
v1f and v2f are the final velocities of the car and truck, respectively.

In this case, we are given:
m1 (mass of car) = 853 kg
m2 (mass of truck) = 1450 kg
v1 (initial velocity of car) = 0 m/s (since the car is stopped)
v2 (initial velocity of truck) = 16.0 m/s

We need to solve for v1f and v2f.

Using the conservation of linear momentum equation, we have:

(853 kg * 0 m/s) + (1450 kg * 16.0 m/s) = (853 kg * v1f) + (1450 kg * v2f)

0 kg m/s + 23200 kg m/s = 853 kg * v1f + 1450 kg * v2f

23200 kg m/s = 853 kg * v1f + 1450 kg * v2f

To solve for v1f and v2f, we need another equation.

Since the collision is considered approximately elastic, that means kinetic energy is conserved. Thus, we can use the equation:

(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2

Substituting the given values:

(1/2) * 853 kg * 0 m/s + (1/2) * 1450 kg * (16.0 m/s)^2 = (1/2) * 853 kg * v1f^2 + (1/2) * 1450 kg * v2f^2

Simplifying the equation:

0 + 116,800 kg m^2/s^2 = (1/2) * 853 kg * v1f^2 + (1/2) * 1450 kg * v2f^2

116,800 kg m^2/s^2 = 1/2 * 853 kg * v1f^2 + (1/2) * 1450 kg * v2f^2

We now have a system of two equations with two variables:
23200 kg m/s = 853 kg * v1f + 1450 kg * v2f
116,800 kg m^2/s^2 = (1/2) * 853 kg * v1f^2 + (1/2) * 1450 kg * v2f^2

To solve this system of equations, you can use algebraic methods such as substitution or elimination. The final velocity values, v1f and v2f, will be the solutions to the system.