Question

four A rallroad car of mass 2.25 x 101 kg moving at 3.50 m's collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s (a) What is the speed of the three coupled cars after the collision? m/s (b) How much kinetic energy is lost in the collision? Need Help? ce, an 0.5-kg actor swings from a R-4s-m-long cable that is horizontal when he starts. At the bottom of his are, he picks up his 55.0-kg costar in an In a inelastic collision. What maximum height do they reach after their upward swing? differs from the correct answer by more than 10%. Double check your calculations. m Need Help?

Answer 1

The momentum of railroad car =

The energy of railroad car =

The momentum of two coupled railroad cars =

The kinetic energy of two coupled railroad cars =

Say the final velocity of the system = v₃

So, according to the conservation of momentum:

The energy of the final system =

So, loss of energy:

2)

At the starting point his potential energy relative to the bottom part of swing =

, m is mass, g gravitational acceleration of earth and h is the height from the bottom part.

Say, at the bottom point his kinetic energy is v. His kinetic energy will be then:

The energy stays conserved, so,

Total mass at the bottom = m+55 kg

After the inelastic collision, the speed becomes v1 (say)

according to the conservation of momentum:

so total energy becomes:

Say H is the maximum height he reaches, as energy stays conserved,