A missile is fired form the back of a stationary truck. The missile takes off at a speed of 41.7 m/s and has a mass of 53.8 kg. If the truck has a mass of 856 kg, what is the recoil speed of the truck?
If you fired it straight up, the truck would not move.
If you fired it straight back and there were no brakes on the truck then
53.8 (41.7) = (856-53.8) v
To find the recoil speed of the truck, we can use the law of conservation of momentum. According to this law, the total momentum before the missile is fired will be equal to the total momentum after the missile is fired. The momentum of an object is calculated by multiplying its mass by its velocity.
Let's assume that the initial velocity of the truck is 0 m/s.
The total momentum before the missile is fired is the sum of the momentum of the truck and the momentum of the missile. After the missile is fired, the truck will move in the opposite direction with a certain velocity, and the missile will move in its own direction with its given velocity.
Before the missile is fired:
Momentum of the truck (mt) = mass of the truck (mt) × initial velocity of the truck (vt)
Momentum of the missile (mm) = mass of the missile (mm) × initial velocity of the missile (vm)
After the missile is fired:
Momentum of the truck (mt') = mass of the truck (mt) × final velocity of the truck (vt')
Momentum of the missile (mm') = mass of the missile (mm) × final velocity of the missile (vm')
According to the law of conservation of momentum:
Total momentum before = Total momentum after
(mt + mm) = (mt' + mm')
Since the initial velocity of the truck is 0, the term "mt" on the left side of the equation can be ignored.
(mm) = (mt' + mm')
(mm) = (mt) × (vt')
(mm) = (856 kg) × (vt') [Substituting the given mass of the truck]
Now, let's use the given values to solve for the final velocity of the truck:
(mm) = (856 kg) × (vt')
(53.8 kg × 41.7 m/s) = (856 kg) × (vt')
Solving for vt':
(53.8 kg × 41.7 m/s) / (856 kg) = (vt')
(2239.46 kg·m/s) / (856 kg) = (vt')
2.6184 m/s = (vt')
Therefore, the recoil speed of the truck is approximately 2.6184 m/s in the opposite direction.