A man is in a boat out away from shore. Th mass of the man, the boat and everything in it is 210 kg. The boat is not moving. The man throws and anchor with a mass of 26 kg straight out from the boat with a speed of 6.6 m/s. What is the resulting speed of the boat?
This is really hard
To find the resulting speed of the boat, we can apply the law of conservation of momentum. According to this law, the total momentum before the anchor is thrown equals the total momentum after the anchor is thrown.
The total momentum before the anchor is thrown is equal to the momentum of the man, the boat, and everything in it. Since the boat is not moving initially, this momentum is zero.
The total momentum after the anchor is thrown is the sum of the momentum of the man, the boat, and everything in it, and the momentum of the anchor. The momentum of an object is given by the product of its mass and velocity.
Let's denote the velocity of the man and the boat after the anchor is thrown as V. The momentum of the man, the boat, and everything in it after the anchor is thrown is equal to (210 kg) * V. The momentum of the anchor is equal to (26 kg) * (6.6 m/s).
According to the law of conservation of momentum, the total momentum before the anchor is thrown is equal to the total momentum after the anchor is thrown:
0 = (210 kg) * V + (26 kg) * (6.6 m/s)
To find the resulting speed of the boat (V), we can rearrange the equation:
(210 kg) * V = -(26 kg) * (6.6 m/s)
V = -((26 kg) * (6.6 m/s)) / (210 kg)
Simplifying the equation gives us:
V ≈ -0.82 m/s
Therefore, the resulting speed of the boat after the man throws the anchor is approximately 0.82 m/s in the opposite direction of the thrown anchor.