A two-stage rocket moves in space at a constant velocity of +4890 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1380-kg upper stage is +5660 m/s. What is the velocity (magnitude and direction) of the 2970-kg lower stage immediately after the explosion?

M1*V + M2*V = M1*V1 + M2*V2.

1380*4890 + 2970*4890 = 1380*5660 + 2970*V2. V2 = ?.

To find the velocity (magnitude and direction) of the lower stage immediately after the explosion, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.

The momentum of an object is defined as the product of its mass and velocity. Therefore, we can write the equation:

(mass1 * velocity1) + (mass2 * velocity2) = (mass1 * velocity1') + (mass2 * velocity2')

Where:
mass1 = mass of the upper stage = 1380 kg
velocity1 = velocity of the upper stage before the explosion = +4890 m/s
mass2 = mass of the lower stage = 2970 kg
velocity2 = velocity of the lower stage before the explosion (which is zero as it's still attached)
velocity1' = velocity of the upper stage after the explosion = +5660 m/s
velocity2' = velocity of the lower stage after the explosion (what we need to find)

Let's substitute the given values into the equation:

(1380 kg * 4890 m/s) + (2970 kg * 0 m/s) = (1380 kg * 5660 m/s) + (2970 kg * velocity2')

Now, we solve for velocity2':

(1380 kg * 4890 m/s) + (2970 kg * 0 m/s) = (1380 kg * 5660 m/s) + (2970 kg * velocity2')
6742200 kg·m/s = 7864800 kg·m/s + 2970 kg * velocity2'

Rearranging the equation to solve for velocity2':

2970 kg * velocity2' = 7864800 kg·m/s - 6742200 kg·m/s
2970 kg * velocity2' = 1122600 kg·m/s

Dividing both sides of the equation by 2970 kg:

velocity2' = 1122600 kg·m/s / 2970 kg
velocity2' ≈ 378.79 m/s

Therefore, the magnitude of the velocity of the lower stage immediately after the explosion is approximately 378.79 m/s. Since it was initially in the same direction as the upper stage, the direction remains the same.

Therefore, the velocity (magnitude and direction) of the lower stage immediately after the explosion is approximately +378.79 m/s (in the same direction as the upper stage).