A two-stage rocket moves in space at a constant velocity of +4660 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1240-kg upper stage is +5570 m/s. What is the velocity (magnitude and direction) of the 2080-kg lower stage immediately after the explosion?

Question

A two-stage rocket moves in space at a constant velocity of +4660 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1240-kg upper stage is +5570 m/s. What is the velocity (magnitude and direction) of the 2080-kg lower stage immediately after the explosion?

Answer 1

initial momentum=final momentum

(2080+1240)4660=2080*V+1240*5570
solve for V

Answer 2

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 the principle of conservation of momentum, the total momentum before the explosion should be equal to the total momentum after the explosion.

The momentum of an object is given by the product of its mass and velocity. So, we can calculate the momentum of each stage before and after the explosion.

Let's denote the velocity of the lower stage after the explosion as V_lower and the velocity of the upper stage after the explosion as V_upper.

Given:
Mass of the upper stage (m_upper) = 1240 kg
Velocity of the upper stage before the explosion (V_upper,before) = +4660 m/s
Velocity of the upper stage after the explosion (V_upper,after) = +5570 m/s

Mass of the lower stage (m_lower) = 2080 kg

Before the explosion, the total momentum of the system is:
Total momentum before = (m_upper * V_upper,before) + (m_lower * 0) [The lower stage is stationary]

After the explosion, the total momentum of the system is:
Total momentum after = (m_upper * V_upper,after) + (m_lower * V_lower)

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

(m_upper * V_upper,before) = (m_upper * V_upper,after) + (m_lower * V_lower)

Now, let's solve for V_lower:

V_lower = (m_upper * V_upper,before - m_upper * V_upper,after) / m_lower

Plugging in the given values, we have:

V_lower = (1240 kg * 4660 m/s - 1240 kg * 5570 m/s) / 2080 kg

Simplifying the equation:

V_lower = (1240 kg * (4660 m/s - 5570 m/s)) / 2080 kg

V_lower = (1240 kg * (-910 m/s)) / 2080 kg

V_lower = -658 m/s

Therefore, the velocity (magnitude and direction) of the lower stage immediately after the explosion is 658 m/s in the opposite direction (negative sign indicates opposite direction) to the initial movement of the rocket.