A submarine of mass 2.55 × 106 kg and initially at rest fires a torpedo of mass 254 kg. The torpedo has an initial speed of 100 m/s. What is the initial recoil speed of the submarine? Neglect the drag force of the water.
conserving momentum (Initial is zero):
2.55*10^6 v + 254 (100) = 0
v is negative because if it fires out the bow, the sub moves backwards.
To find the initial recoil speed of the submarine, we can use the principle of conservation of momentum. According to this principle, the total momentum before the torpedo is fired is equal to the total momentum after the torpedo is fired.
Before firing the torpedo, the submarine is at rest, so its initial momentum is zero (since momentum is mass times velocity). The initial momentum of the torpedo is given by the product of its mass (254 kg) and its initial speed (100 m/s).
Now, let's denote the initial recoil speed of the submarine as V. After the torpedo is fired, the submarine and the torpedo move in opposite directions. So the momentum of the submarine will be the mass of the submarine (2.55 × 10^6 kg) times the negative recoil speed (-V), and the momentum of the torpedo will be its mass of (254 kg) times its speed (-100 m/s).
Using the principle of conservation of momentum, we can write the equation:
0 + 254 kg * 100 m/s = (2.55 × 10^6 kg) * (-V) + 254 kg * (-100 m/s)
Simplifying the equation, we have:
25,400 kg·m/s = -(2.55 × 10^6 kg) * V - 25,400 kg·m/s
Now, let's solve for V by rearranging the equation:
25,400 kg·m/s + 25,400 kg·m/s = -(2.55 × 10^6 kg) * V
50,800 kg·m/s = -(2.55 × 10^6 kg) * V
V = -50,800 kg·m/s / (-2.55 × 10^6 kg)
V = 19.92 m/s
So, the initial recoil speed of the submarine is approximately 19.92 m/s in the opposite direction to the fired torpedo.