A missile of mass 110 kg is fired from a plane of mass 5500kg initially moving at a speed of 315 m/s. If the speed of the missile relative to the plane is 1060 m/s, what is the final velocity of the plane?
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant before and after an event.
The total momentum of the system (plane + missile) before firing the missile is given by the sum of the momenta of the plane and the missile. After firing the missile, the momentum of the plane and the missile will be in opposite directions since the missile is being ejected.
The momentum of an object is given by the product of its mass and velocity. Let's assign variables to the different quantities:
Mass of the missile (m1) = 110 kg
Mass of the plane (m2) = 5500 kg
Initial velocity of the plane (v2 initial) = 315 m/s
Relative velocity between the missile and the plane (v1 relative) = 1060 m/s
Using the conservation of momentum, we can write the equation:
(m1 + m2) * v initial = m1 * v1 + m2 * v2
where v initial is the initial velocity of the system (plane + missile) before firing the missile, v1 is the final velocity of the missile after firing, and v2 is the final velocity of the plane after firing the missile.
Substituting the given values into the equation, we have:
(110 kg + 5500 kg) * 315 m/s = 110 kg * 1060 m/s + 5500 kg * v2
Simplifying the equation:
(5610 kg) * 315 m/s = (116600 kg*m/s) + (5500 kg) * v2
1767150 kg*m/s = 116600 kg*m/s + 5500 kg * v2
Rearranging the equation to solve for v2, we get:
5500 kg * v2 = 1767150 kg*m/s - 116600 kg*m/s
5500 kg * v2 = 1640550 kg*m/s
Dividing both sides by 5500 kg:
v2 = 298.28 m/s
Therefore, the final velocity of the plane is approximately 298.28 m/s.