A man of mass 70 kg and a boy of mass 30 kg dive of the end of a boat, of mass 150 kg with a horizontal velocity of 2 m/s relativeto the boat. if initially the boat is at rest, findits velocity just after (1) both the man and boy dive off simultaneously (2) the man dives of first followed by the boy
Conserve momentum. (Initial = final)
(1) Man and boy simultaneously jump and push the boat back.
m1 = mass of man
m2 = mass of boy
v1 = velocity of man (and boy)
m3 = mass of boat
v3 =
0 = (m1+m2) v1 + m3 v3
Rearrange:
v3 = - (m1+m2) v1 / m3
= - 100 * 2 / 150
= 1.33 m/s
(2) First, the man pushes back the boat and the boy, then the boy further pushes back the boat.
Key point: When the boy jumps off at v1 _relative_ to the boat, the boat is not at rest.
m1 = mass of man
m2 = mass of boy
v1 = velocity of man
v2 = intermediate velocity of boat
m3 = mass of boat
v4 = final velocity of boat
First jump:
0 = m1 v1 + (m2+m3) v2
Second jump:
(m2+m3) v2 = m2 (v2+v1) + m3 v4
Rearrange:
v2 = - m1 v1 / (m2+m3)
v4 = v2 - (m2 v1)/ m3
Substitute first into the second
v4 = - (m1/(m2+m3)+m2/m3) v1
To solve this problem, we can use the principle of conservation of momentum.
1) When both the man and boy dive off simultaneously, we can consider the system as consisting of the man, boy, and the boat. The initial momentum of the system is zero since the boat is at rest. Since momentum is conserved in the absence of external forces, the final momentum of the system must also be zero.
Let's assume the velocity of the boat just after the dive is v.
The man's momentum is given by: momentum of the man = mass of the man x velocity of the man
= 70 kg x 2 m/s = 140 kg.m/s
The boy's momentum is given by: momentum of the boy = mass of the boy x velocity of the boy
= 30 kg x 2 m/s = 60 kg.m/s
The boat's momentum is given by: momentum of the boat = mass of the boat x velocity of the boat
= 150 kg x v
Since the total momentum before the dive is zero, the total momentum after the dive must also be zero. Therefore, we have the equation:
momentum of the man + momentum of the boy + momentum of the boat = 0
140 kg.m/s + 60 kg.m/s + 150 kg x v = 0
Simplifying the equation, we get:
200 kg.m/s + 150 kg x v = 0
150 kg x v = -200 kg.m/s
v = -200 kg.m/s / 150 kg
v ≈ -1.33 m/s
So, the velocity of the boat just after both the man and boy dive off simultaneously is approximately -1.33 m/s in the opposite direction of the initial velocity.
2) When the man dives off first followed by the boy, the momentum of the system is still conserved. However, we need to consider them as separate events. First, let's calculate the velocity of the boat just after the man dives off.
Using the same method as before, the initial momentum of the system is zero. The momentum of the man is 140 kg.m/s, and the momentum of the boat is 150 kg x v.
Therefore, 140 kg.m/s + 150 kg x v = 0
150 kg x v = -140 kg.m/s
v = -140 kg.m/s / 150 kg
v ≈ -0.933 m/s
So, the velocity of the boat just after the man dives off is approximately -0.933 m/s in the opposite direction of the initial velocity.
Next, when the boy dives off, we need to consider the momentum of the boy and the boat. The boy's momentum is 60 kg.m/s, and the momentum of the boat is 150 kg x v.
Therefore, 60 kg.m/s + 150 kg x v = 0
150 kg x v = -60 kg.m/s
v = -60 kg.m/s / 150 kg
v = -0.4 m/s
So, the velocity of the boat just after the boy dives off is approximately -0.4 m/s in the opposite direction of the initial velocity.
To solve this problem, we need to apply the law of conservation of momentum. The total momentum before and after the dive should be the same.
1. When both the man and boy dive off simultaneously:
Let's assume the velocity of the boat just after the dive is v1, and the velocities of the man and boy are v2 and v3, respectively.
Before the dive, the total momentum is given by the sum of individual momenta:
Initial momentum = (mass of man × velocity of man) + (mass of boy × velocity of boy) + (mass of boat × velocity of boat)
Initial momentum = (70 kg × 0 m/s) + (30 kg × 0 m/s) + (150 kg × 0 m/s)
Initial momentum = 0 kg⋅m/s
After the dive, the total momentum is given by:
Final momentum = (mass of man × velocity of man) + (mass of boy × velocity of boy) + (mass of boat × velocity of boat)
Final momentum = (70 kg × v2) + (30 kg × v3) + (150 kg × v1)
According to the conservation of momentum, the initial and final momentum should be equal:
Initial momentum = Final momentum
0 = (70 kg × v2) + (30 kg × v3) + (150 kg × v1)
Since the boat is at rest initially, the velocity of the boat before the dive is 0 m/s:
v1 = 0 m/s
So, the equation becomes:
0 = (70 kg × v2) + (30 kg × v3)
To find the velocity of the boat just after both the man and boy dive off simultaneously, we need to find v2 and v3. However, we cannot determine their individual velocities only from the given information. Additional information or equations are needed.
2. When the man dives off first followed by the boy:
In this case, the velocity of the boat just after the man dives off (v1) will be different from 0 m/s.
Again, we assume the velocities of the man and boy are v2 and v3, respectively.
Before the dive, the initial momentum is still 0 kg⋅m/s because the boat is initially at rest.
After the dive, the final momentum is given by:
Final momentum = (mass of man × velocity of man) + (mass of boy × velocity of boy) + (mass of boat × velocity of boat)
Final momentum = (70 kg × v2) + (30 kg × v3) + (150 kg × v1)
Applying conservation of momentum, the equation becomes:
0 = (70 kg × v2) + (30 kg × v3) + (150 kg × v1)
To find the velocity of the boat just after the man dives off first, followed by the boy, we need to know the specific values of v2, v3, and v1. Without additional information or equations, these specific velocities cannot be determined.
In both cases, we need additional information to solve for the velocities and calculate the boat's velocity just after the dives.