A two staged rocket moves in space at a constant velocity of 4900m/s. The two stages are then separated by a small explosive charge between them. Immediately after the explosion the velocity of the 1200kg upper stage is 5700m/s in the same direction as before the explosion. What is the velocity (magnitude and direction) of he 2400kg lower stage after the explosion.

conservation of momentum applies

you are given both masses.

(1200+2400)4900=1200*5700+2400*V

To solve the problem, we'll use the principle of conservation of momentum. According to this principle, the total momentum before the explosion should be equal to the total momentum after the explosion.

Let's start by finding the momentum of the upper stage (1200kg) before and after the explosion.

1. Before explosion:
Momentum of the upper stage before the explosion = mass × velocity
Momentum (before) = 1200 kg × 4900 m/s

2. After explosion:
Momentum of the upper stage after the explosion = mass × velocity
Momentum (after) = 1200 kg × 5700 m/s

Since the momentum is conserved, we can equate the momentum before and after the explosion:

1200 kg × 4900 m/s = 1200 kg × 5700 m/s

Now, let's find the momentum of the lower stage (2400kg) after the explosion.

Total momentum after the explosion = momentum of upper stage + momentum of lower stage
Momentum (after) = 1200 kg × 5700 m/s + mass of lower stage × velocity of lower stage

We can rewrite the equation using the values we know:

1200 kg × 5700 m/s = 1200 kg × 5700 m/s + 2400 kg × velocity of lower stage

Simplifying the equation:

0 = 2400 kg × velocity of lower stage

Therefore, the velocity of the lower stage after the explosion is 0 m/s. This means that the lower stage stops moving after the separation.

To solve this problem, we need to apply the law of conservation of momentum, which states that the total momentum of a system remains constant if no external forces are acting on it.

Given that the velocity of the two-stage rocket before the explosion is 4900 m/s and the mass of the upper stage is 1200 kg, we can calculate the momentum of the upper stage before the explosion:

Momentum before explosion = mass × velocity
P1 = 1200 kg × 4900 m/s
P1 = 5880000 kg·m/s

After the explosion, the upper stage separates from the lower stage. The momentum of the upper stage will now be given by its new mass (1200 kg) and new velocity (5700 m/s):

P2 (upper stage) = 1200 kg × 5700 m/s
P2 = 6840000 kg·m/s

Since momentum is conserved, the total momentum after the explosion will be equal to the momentum before the explosion:

Total momentum after explosion = P2 (upper stage) + P3 (lower stage)

Now we can solve for the momentum of the lower stage. We know that the total mass of the rocket is 1200 kg + 2400 kg = 3600 kg. Let's denote the velocity of the lower stage after the explosion as v3:

Momentum after explosion = 6840000 kg·m/s + (2400 kg × v3) kg·m/s

Since the total momentum after the explosion is equal to the momentum before the explosion, we can set up the equation:

5880000 kg·m/s = 6840000 kg·m/s + (2400 kg × v3) kg·m/s

Simplifying the equation:

-960000 kg·m/s = 2400 kg × v3

Dividing both sides by 2400 kg gives us:

v3 = -400 m/s

So, the velocity of the lower stage after the explosion is -400 m/s in the opposite direction of the original velocity.