A 300kg motorboat is turned off as it approaches a dock and it coasts in toward the dock at .5 m/s. Isaac, whose mass is 62kg, jumps off the front of the boat with a speed of 3m/s relative to the boat. what is the velocity of the boat after Isaac jumps?
The combined momentum of motorboat and Isaac remains the same. Choose a coordinate system based on land. The momentum conservation equation can then be written
0.5 m/s*(Mb + Mi) = Mb*Vb + Mi*(Vi+0.5)
Solve for the unknown velocity of the boat afterwards, Vb.
0.5*362 = 300 Vb + 62*3.5
300 Vb = 181 - 217 = -36
Vb = -0.12 m/s (backwards)
A 300 kg motorboat is turned off as it approaches a dock and it coasts in toward the dock at 6.0 m/s. A lone passenger named Isaac, whose mass is 62.0 kg, jumps off the front of the boat with a speed of 12.0 m/s. What is the velocity of the boat after Isaac jumps?
Well, if Isaac jumps off the boat with a speed of 3 m/s relative to the boat, we can consider him as a separate system from the motorboat.
Since there is no external force acting on the system (boat + Isaac) after Isaac jumps off, the total momentum of the system must be conserved.
Before Isaac jumps off, the total momentum of the system is given by:
Total Momentum = (mass of boat + mass of Isaac) x velocity of boat
After Isaac jumps off, the boat starts to move in the opposite direction with a new velocity, let's call it V'. The total momentum of the system is still conserved:
Total Momentum = (mass of boat + mass of Isaac) x V'
Since the initial momentum is equal to the final momentum, we can set up an equation:
(mass of boat + mass of Isaac) x velocity of boat = (mass of boat + mass of Isaac) x V'
We can cancel out the mass of boat + mass of Isaac on both sides of the equation:
velocity of boat = V'
So, the velocity of the boat after Isaac jumps is equal to the speed at which Isaac jumped off the boat, which is 3 m/s.
To solve this problem, we can apply the principle of conservation of linear momentum. According to this principle, the total momentum of an isolated system remains constant before and after an event (such as Isaac jumping off the boat). In other words, the momentum of the boat before Isaac jumps must equal the momentum of the boat and Isaac combined after he jumps.
To find the velocity of the boat after Isaac jumps, we need to calculate its momentum before and after the event.
Momentum (p) is defined as the product of an object's mass (m) and its velocity (v).
Before Isaac jumps:
Momentum of the boat before = mass of the boat (m₁) × velocity of the boat before (v₁)
= 300 kg × 0.5 m/s
After Isaac jumps:
Momentum of the boat and Isaac after = (mass of the boat (m₁) + mass of Isaac (m₂)) × velocity of the boat and Isaac after
To find the velocity of the boat after Isaac jumps, we can rearrange the equation and solve for v₂:
v₂ = (momentum of the boat before - momentum of Isaac after) / mass of the boat
First, let's calculate the momentum of Isaac after he jumps:
Momentum of Isaac after = mass of Isaac (m₂) × velocity of Isaac (vᵢ)
= 62 kg × 3 m/s
Now we can substitute the values into the formula to calculate the velocity of the boat after Isaac jumps:
v₂ = (300 kg × 0.5 m/s - 62 kg × 3 m/s) / 300 kg
Simplifying the equation:
v₂ = (150 kg·m/s - 186 kg·m/s) / 300 kg
v₂ = -36 kg·m/s / 300 kg
v₂ = -0.12 m/s
Therefore, the velocity of the boat after Isaac jumps is -0.12 m/s. The negative sign indicates that the boat is moving in the opposite direction to its original motion.
boat is m1 = 300 kg
Isaac is m2 = 62 kg
towards dock is +
away from dock is -
original velocity of both is
v = +0.50 m/s
v2' = final velocity of Isaac
v1' = final velocity of boat
v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
(m1+m2)v = m1v1' + m2v2'
v1' = 0.014 m/s away from the dock (boat) and v2' - 2.986 m/s towards the dock (Isaac)
boat is m1 = 300 kg
Isaac is m2 = 62 kg
towards dock is +
away from dock is -
original velocity of both is
v = +0.50 m/s
v2' = final velocity of Isaac
v1' = final velocity of boat
v2'-v1' = +3.0 m/s (because Isaac jumps off the front of the boat & is moving toward the dock; this will make v2' positive and v1' negative.
(m1+m2)v = m1v1' + m2v2'
v1' = 0.014 m/s away from the dock (boat) and v2' = 2.986 m/s towards the dock (Isaac)