Solve Velocity Problem: Boat w/ 2 Boys Jumping

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The discussion centers on calculating the velocity of a boat after two boys jump off in opposite directions. The first boy, weighing 65 kg, jumps off the front at 2 m/s, while the second boy, weighing 68 kg, jumps off the back at 4 m/s. The initial velocity of the boat is 3 m/s, and its mass is 100 kg. The calculations involve determining the momentum of each boy and the boat's resulting velocity after the jumps. The final velocity calculation appears to be incorrect, as the momentum conservation principle must be applied correctly to account for the boat's mass.
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a boat is moving at 3 meters/sec one boy wieghs 65 kgjumps off the front at 2 m/s. the second boy jumps out of the back of the boat, he wieghs 68kg and jumps at a speed of 4 m/s. if the boats mass is 100kg. what is the velocity of the boat.


what i did was multiply 65 times 2 to get the force that the 1st bot jumped of the boat a. i got 130kg m/s. then i did the same for the second boy but i used 68 and 4 and got 272kg m/s. i subracted 130 from 272 and got 142. after that i added up the masses of the two boys and got 133. i divided 142 by 133 and got 1.06 i added the initial 3 m/s and got 4.06. Is this right? it seemed too easy.
 
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(mass X velocity) is momentum, not force.

Did they give you the mass of the boat?
 
no it wasn't given
 
ooops the boat was 100 kg.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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