An 85.0 kg fisherman decides to jump from his boat onto the dock. If the velocity of the fisherman is 4.30 m/s to the
west and the boat is 185 kg, what is the velocity of the boat after the fisherman jumps out?
conservation of momentum applies
Massman*velocityman+massboat*velocityboat=0 Remember, the initial momentum of both was zero.
To find the velocity of the boat after the fisherman jumps out, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jump must be equal to the total momentum after the jump.
The momentum of an object can be calculated by multiplying its mass by its velocity. In this case, let's assume the positive direction is to the west.
The initial momentum of the system (fisherman + boat) before the jump is given by:
Initial momentum = (mass of fisherman × velocity of fisherman) + (mass of boat × velocity of boat)
Given:
Mass of fisherman = 85.0 kg
Velocity of fisherman = 4.30 m/s to the west
Mass of boat = 185 kg
So, the initial momentum is:
Initial momentum = (85.0 kg × (-4.30 m/s)) + (185 kg × velocity of boat)
Now, after the fisherman jumps out, the only momentum remaining is that of the boat. The momentum of the boat after the jump can be calculated by multiplying its mass by its velocity.
Final momentum = mass of boat × velocity of boat
Since momentum is conserved, we can equate the initial and final momentum:
(85.0 kg × (-4.30 m/s)) + (185 kg × velocity of boat) = 185 kg × velocity of boat
Simplifying the equation, we can solve for the velocity of the boat:
(-365.5 kg·m/s) + (185 kg × velocity of boat) = 185 kg × velocity of boat
(-365.5 kg·m/s) = (-365 kg × velocity of boat)
Dividing both sides by 365 kg, we get:
(-365.5 kg·m/s) / 365 kg = velocity of boat
After simplification, the velocity of the boat after the fisherman jumps out is approximately equal to the initial velocity of the fisherman, which is 4.30 m/s to the west.