An 88kg fisherman jumps from a dock into a 131kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.7m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat. East is positive.
To find the momentum of the fisherman as he jumps, we use mass times velocity (mv). This leads us to 88kg x 4.7 m/s, giving us 413.6 kg m/s. Since this is how much momentum we start with, it will also be how much we end with. We still use mass times velocity (mv), but this time the boat is included in the mass (m). 88 kg + 131 kg = 219 kg, so now we solve for the velocity: 219 kg x v = 413.6 kg m/s. After dividing 413.6 kg m/s by 219 kg, we get the final velocity of 1.88 m/s. Since this is to the west, this value will be negative, giving us a FINAL ANSWER of -1.88 m/s.
Apply conservation of momentum. If east is positive, the final momentum and velocity will be negative.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after the interaction remains the same.
The momentum (p) of an object is calculated using the formula p = m * v, where m is the mass of the object and v is its velocity.
Let's denote the final velocity of the fisherman and the boat as V. Since the rowboat is initially at rest, its initial velocity is 0 m/s.
The momentum before the jump is given by:
momentum_before = (mass_fisherman * velocity_fisherman) + (mass_boat * velocity_boat)
= (88 kg * 4.7 m/s) + (131 kg * 0 m/s)
= 413.6 kg·m/s
The momentum after the jump is given by:
momentum_after = (mass_fisherman + mass_boat) * V
According to the conservation of momentum, the initial momentum is equal to the final momentum:
momentum_before = momentum_after
Now we can solve for V:
413.6 kg·m/s = (88 kg + 131 kg) * V
413.6 kg·m/s = 219 kg * V
V = 413.6 kg·m/s / 219 kg
V ≈ 1.886 m/s
Therefore, the final velocity of the fisherman and the boat is approximately 1.886 m/s to the west.
To find the final velocity of the fisherman and the boat, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jump is equal to the total momentum after the jump.
Momentum is calculated by multiplying mass and velocity. Let's assume the final velocity of the fisherman and the boat is Vf, and the initial velocity of the boat is Vi (which is 0 since it is at rest).
Before the jump:
The mass of the fisherman is 88 kg, and his velocity is -4.7 m/s (west direction).
The mass of the boat is 131 kg, and its velocity is 0 m/s since it is at rest.
The total momentum before the jump is given by:
Total momentum before = (mass of the fisherman * velocity of the fisherman) + (mass of the boat * velocity of the boat)
Total momentum before = (88 kg * -4.7 m/s) + (131 kg * 0 m/s) = -412.4 kg·m/s
After the jump:
The mass of the fisherman and the boat combined is the sum of their individual masses: 88 kg + 131 kg = 219 kg.
The velocity of the fisherman and the boat combined is Vf (to the east) since the fisherman jumps to the boat.
The total momentum after the jump is given by:
Total momentum after = (mass of the fisherman and boat * velocity of the fisherman and boat)
Total momentum after = (219 kg * Vf)
According to the conservation of momentum principle, the total momentum before the jump is equal to the total momentum after the jump:
-412.4 kg·m/s = 219 kg * Vf
Now, let's solve this equation for Vf:
Vf = (-412.4 kg·m/s) / 219 kg
Vf ≈ -1.88 m/s
The final velocity of the fisherman and the boat is approximately -1.88 m/s to the east.