A 70kg person dives horizontally from a 200kg boat with a speed of 2 m/s. What is the recoil speed of the boat?
To find the recoil speed of the boat, we can use the principle of conservation of momentum. When the person jumps off the boat, the total momentum before and after the jump should be the same.
Before the jump, the momentum is given by the formula:
Total momentum = momentum of the person + momentum of the boat
The momentum of an object is calculated as the product of its mass and velocity:
Momentum = mass × velocity
Let's assume that the boat's recoil speed after the jump is denoted by Vb. Since the person jumps horizontally, their momentum is entirely in the horizontal direction. Therefore, we can ignore the vertical components of the velocities.
Given:
Mass of the person (mp) = 70 kg
Mass of the boat (mb) = 200 kg
Initial velocity of the person (vp) = 2 m/s
Using the conservation of momentum, the total momentum before and after the jump is equal. So, we can write the equation as:
(70 kg × 2 m/s) + (200 kg × 0 m/s) = 0 kg × Vb + 200 kg × Vb
Simplifying the equation:
140 kg·m/s = 200 kg × Vb
Dividing both sides by 200 kg:
0.7 m/s = Vb
Therefore, the recoil speed of the boat is 0.7 m/s.