A two-stage rocket moves in space at a constant velocity of 4800 m/s. The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1290-kg upper stage is 5600 m/s in the same direction as before the explosion. What is the velocity (magnitude and direction) of the 2300-kg lower stage after the explosion?
conservation of momentum (again)
(2300 + 1290) * 4800 =
(1290 * 5600) + (2300 * v)
To solve this problem, we need to apply the law of conservation of momentum. According to this law, 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. Let's denote the mass of the upper stage as m1 (1290 kg) and its velocity before the explosion as v1 (4800 m/s). The mass of the lower stage will be denoted as m2 (2300 kg), and we need to find its velocity after the explosion, which we'll denote as v2.
Before the explosion, the total momentum is given by:
Total momentum before = (mass of upper stage) * (velocity of upper stage) + (mass of lower stage) * (velocity of lower stage)
= m1 * v1 + m2 * 0 (lower stage is stationary)
After the explosion, the total momentum is given by:
Total momentum after = (mass of upper stage) * (velocity of upper stage after explosion) + (mass of lower stage) * (velocity of lower stage after explosion)
= m1 * v2 + m2 * v2 (both stages move together with the same velocity v2)
Now, equating the total momentum before and after the explosion:
m1 * v1 + m2 * 0 = m1 * v2 + m2 * v2
Simplifying the equation:
m1 * v1 = (m1 + m2) * v2
Now, plug in the given values:
(1290 kg) * (4800 m/s) = (1290 kg + 2300 kg) * v2
Solving for v2:
(1290 kg * 4800 m/s) / (1290 kg + 2300 kg) = v2
v2 = 5600 m/s
Therefore, the velocity of the lower stage after the explosion is 5600 m/s in the same direction as before the explosion.
To find the velocity of the lower stage after the explosion, we can 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.
First, let's calculate the initial momentum before the explosion. The initial momentum is the product of the mass and velocity of the upper and lower stages combined:
Initial momentum before the explosion = (Mass of upper stage × Velocity of upper stage) + (Mass of lower stage × Velocity of lower stage)
Given:
Mass of upper stage = 1290 kg
Velocity of upper stage before the explosion = 4800 m/s
Mass of lower stage = 2300 kg
Let's calculate the initial momentum before the explosion:
Initial momentum before the explosion = (1290 kg × 4800 m/s) + (2300 kg × 4800 m/s)
= (6192000 kg·m/s) + (11040000 kg·m/s)
= 17232000 kg·m/s
Now, let's calculate the final momentum after the explosion. The final momentum is the product of the mass and velocity of the upper and lower stages after the explosion:
Final momentum after the explosion = (Mass of upper stage × Velocity of upper stage after the explosion) + (Mass of lower stage × Velocity of lower stage after the explosion)
Given:
Velocity of upper stage after the explosion = 5600 m/s
Let's calculate the final momentum after the explosion:
Final momentum after the explosion = (1290 kg × 5600 m/s) + (2300 kg × Velocity of lower stage after the explosion)
We know that the total momentum before the explosion is equal to the total momentum after the explosion. Therefore,
Initial momentum before the explosion = Final momentum after the explosion
17232000 kg·m/s = (1290 kg × 5600 m/s) + (2300 kg × Velocity of lower stage after the explosion)
Now, let's solve for the velocity of the lower stage after the explosion:
(2300 kg × Velocity of lower stage after the explosion) = 17232000 kg·m/s - (1290 kg × 5600 m/s)
Velocity of lower stage after the explosion = (17232000 kg·m/s - (1290 kg × 5600 m/s)) / 2300 kg
Calculating the value of the right-hand side:
Velocity of lower stage after the explosion = (17232000 kg·m/s - 7240000 kg·m/s) / 2300 kg
= 9992000 kg·m/s / 2300 kg
≈ 4340.87 m/s
Therefore, the velocity of the 2300-kg lower stage after the explosion is approximately 4340.87 m/s in the same direction as before the explosion.