Let the mass of 2500 kg car be and it's velocity be
and the mass of 1500 kg car be
and it's velocity be
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Answer:
18.2145 meters
Explanation:
Using the conservation of momentum, we have that:
m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:
The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.
If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is: