A man mass 75 kg and a boy mass 25 kg dives off the end of a boat of mass 300 kg so that their relative horizontal velocity with respect to the boat is 3 m/s. if initially the boat is at rest find its final velocity if (1) the two dive off simultaneously (2) the man dives first followed by the boy
Second Problem:
Momentum is still conserved on each leap (of course) but note that when the boy jumps off at 3m/s _relative_ to the boat, the boat is no longer at rest.
M1 = 75 kg mass of man
M2 = 25 kg mass of boy
M3 = 300 kg mass of boat
V1 = 3 m/s launch velocity (relative to boat)
V2 = ? velocity of boat (and boy) after man leaps.
V3 = ? velocity of boat after boy leaps.
First jump:
0 = M1 V1 + (M2+M3) V2
or
V2 = -M1 V1 / (M2+M3)
Second jump:
(M2+M3) V2 = M2 (V2+V1) + M3 V3
or
V3 = V2 - M2 V1 / M3
Substituting from the first jump:
V3 = -M1 V1/(M2+M3) - M2 V1/M3
Thus using the values:
V3 = -(75)(3)/(75+300) - (25)(3)/(300)
Finally:
V3 = -0.85 m/s
The boat does not move back as fast in this case, because some of the impulse from the man's leap has been imparted to the boy.
I do not know how my answer earlier was reposted, but I think I should have went through the math. There is agreement that the velocity will be 1m/s for the first question. However, there is disagreement concerning the velocity for the second question.
Let
M1=300 kg
V1i=0
V1f=?
M2=25kg
V2i=0
V2f=?
M3=75kg
V3i=0
V3f=-3m/s
***I don't know what direction that each is traveling, but I will let the jumpers' velocity be negative when they jump, since that will imply in the opposite direction.
Momentum Initial=Momentum Final
M1V1i + M2V2i+M3V3i=M1V1f + M2V2f+M3V3f
300kg(0) + 25kg(0) + 75kg(0)=300kg(V1f) + 25kg(V2f) + 75kg(-3m/s)
0=300kg*(V1f) + 25 kg*(V2f) + (75kg)*(-3m/s)
V1f=V2f, so both equal V.
0=300kg*(V) + 25 kg*(V)+(75kg)*(-3m/s)
0=-225J + V(300kg+ 25 kg)
0=-225J + V(325kg)
225J/(325kg)=V
0.692m/s=V
Now, when the boy jumps off, the boat and the boy have a initial velocity of 0.692m/s, but the final velocity of the boy will be -3m/s in the opposite direction.
So let
M1=300 kg
V1i=0.692 m/s
V1f=?
M2=25kg
V2i=0.692 m/s
V2f=3 m/s
M1V1i+M2V2i=M1V1f+M2V2f
V1f=V2f, so both equal V, which is equal to 0.692 m/s
M1V+M2V=M1V1f+M2V2f
Solve for V2f
(300kg*0.692m/s)+ (25kg*0.692m/s)=(300kg)V1f + (25kg*-3m/s)
0.692m/s*(300kg+25kg)=(300kg)V1f + (25kg*-3m/s)
225J=300kg*V1f + (-75J)
225J+75J=300kg*V1f
300J=300kg*V1f
300J/300kg=V1f
1 m/s=V1f
It doesn't make a difference if they jump at the same time or right after each other. The only way that the velocity of the boat will change is if there is a long protracted wait after the initial jump.
(1) If the man and the boy dive off simultaneously, we can use the law of conservation of momentum to calculate the final velocity of the boat.
The initial momentum of the system (boat + man + boy) is zero since the boat is at rest:
Initial momentum = 0
The final momentum of the system should also be zero, as the boat and the people are still connected:
Final momentum = 0
Let's denote the final velocities of the boat, man, and boy as Vb, Vm, and Vb, respectively.
According to the law of conservation of momentum:
(mass of boat * initial velocity of boat) + (mass of man * initial velocity of man) + (mass of boy * initial velocity of boy) = (mass of boat * final velocity of boat) + (mass of man * final velocity of man) + (mass of boy * final velocity of boy)
(300 * 0) + (75 * 0) + (25 * 0) = (300 * Vb) + (75 * 3) + (25 * 3)
0 = 300Vb + 225 + 75
0 = 300Vb + 300
300Vb = -300
Vb = -1
When the man and the boy dive off simultaneously, the boat ends up moving in the opposite direction with a final velocity of -1 m/s.
(2) If the man dives first followed by the boy, we need to consider the individual momenta and then calculate the overall momentum.
For the man diving first:
Initial momentum = 0 (same as before)
Final momentum = (mass of boat * final velocity of boat) + (mass of man * final velocity of man) = (300 * Vb) + (75 * Vm)
For the boy diving after the man:
Initial momentum = (mass of boat * final velocity of boat) + (mass of man * final velocity of man)
Final momentum = (mass of boat * final velocity of boat) + (mass of man * final velocity of man) + (mass of boy * final velocity of boy) = (300 * Vb) + (75 * Vm) + (25 * Vb)
Since the relative horizontal velocity of the man and the boy with respect to the boat is 3 m/s, the final velocity of the boat with respect to the ground will be Vb + Vm. This is because when the man and the boy are on the boat, their relative velocity to the ground is the same as the boat's velocity.
So, the final velocity of the boat when the man dives first followed by the boy will be -1 + Vm.
Note: Without knowing the values of Vm (final velocity of the man) and Vb (final velocity of the boat), we cannot determine the final velocity of the boat.
To find the final velocity of the boat, we can first consider the law of conservation of momentum. According to this law, the total momentum before the dive should be equal to the total momentum after the dive.
(1) When the two dive off simultaneously:
Let's assume the final velocity of the boat after the dive is V. Since the boat is at rest initially, its initial momentum is zero.
The total momentum before the dive is the sum of the momentum of the man, the boy, and the boat:
Initial momentum = 0 + (75 kg × 3 m/s) + (25 kg × 3 m/s) = 0 + 225 kg m/s + 75 kg m/s = 300 kg m/s.
After the dive, the total momentum should still be equal to 300 kg m/s. So we have:
Total momentum = 300 kg m/s = (300 kg + 75 kg + 25 kg) × V
Simplifying the equation, we get:
300 kg m/s = 400 kg × V
Dividing both sides by 400 kg, we find:
V = 300 kg m/s / 400 kg
V = 0.75 m/s
Therefore, the final velocity of the boat when the two dive off simultaneously is 0.75 m/s.
(2) When the man dives first followed by the boy:
In this case, the momentum of the boat changes after the man jumps and before the boy jumps. To find the final velocity of the boat, we need to consider these two separate moments.
First, when the man jumps, we can use the same equation as before to find the velocity of the boat:
300 kg m/s = (300 kg + 75 kg) × V1
V1 = 300 kg m/s / 375 kg
V1 = 0.8 m/s
Now, when the boy jumps, we need to find the resulting change in velocity of the boat. To do this, we can again apply the conservation of momentum:
Final momentum = (300 kg + 25 kg) × V2
The initial momentum after the man's jump is:
Initial momentum = (375 kg × 0.8 m/s)
So, the final momentum should be equal to this initial momentum:
(375 kg × 0.8 m/s) = (325 kg) × V2
V2 = (375 kg × 0.8 m/s) / 325 kg
V2 ≈ 0.923 m/s
Therefore, the final velocity of the boat when the man dives first followed by the boy is approximately 0.923 m/s.
This is a momentum problem and momentum is conserved:
M1V+M2V=M3V
Let
M1=75 kg
M2=25 kg
M3=300 kg
V1=3 m/s
V2= 3 m/s
V3=?
(3m/s)*(75 kg +25 kg)/300 kg=V3
v=1 m/s
Second problem:
Answer Should be the same; momentum is conserved.
But I am not 100% sure.