# collision of railroad cars and gravitational attraction

1. A 9000kg railroad car moving at 3m/s strikes a stationary (not moving) 5000kg railroad car. what is the kinetic energy of the 9000kg railroad car before the collision? What is the momentum of the 9000kg railroad car before the collision? If the two railroad cars couple and move off together after the collision, what is their velocity?

2. A star has a planet held in orbit by gravitational attraction. If the mass of the star were doubled, what effect would this have on the gravitational attraction between the star and its planet ? If the distance between the stars and planet were 3 times as great, what effect would this have on their gravitational attraction for each other?

Â© BrainMass Inc. brainmass.com December 24, 2021, 7:38 pm ad1c9bdddfhttps://brainmass.com/physics/planets/collision-railroad-cars-gravitational-attraction-206190

## SOLUTION This solution is **FREE** courtesy of BrainMass!

1. What is the kinetic energy of the 9000kg railroad car before the collision?

kinetic energy is 0.5mv^2 = 0.5 * 9000 * 3^2 = 40500 J.

What is the momentum of the 9000kg railroad car before the collision?

mv = 9000 * 3 = 27000 kg m/s.

If the two railroad cars couple and move off together after the collision, what is their velocity?

Apply conservation of momentum,

mv = (m + m1)v_f

v_f = 9000 * 3 / (9000 + 5000) = 1.93 m/s

the velocity after the collision is 1.93 m/s.

2. By newton's law of universal gravitation, the gravitational attraction is

F = G* mM/r^2,

where G is the gravitational constant, m is mass of the star, and M is the mass of the planet, r is the distance between them.

When the mass m is doubled, F1 = G * (2m) * M/r^2 = 2F, the attraction is doubled too.

When the distance is doubled, F2 = G * mM/(2r)^2 = F/4, so the gravitational attraction is only one fourth of previous one.

https://brainmass.com/physics/planets/collision-railroad-cars-gravitational-attraction-206190