This is a momentum problem and momentum is conserved:
V2= 3 m/s
(3m/s)*(75 kg +25 kg)/300 kg=V3
Answer Should be the same; momentum is conserved.
But I am not 100% sure.
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.
0 = M1 V1 + (M2+M3) V2
V2 = -M1 V1 / (M2+M3)
(M2+M3) V2 = M2 (V2+V1) + M3 V3
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)
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.
***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)
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.
V1f=V2f, so both equal V, which is equal to 0.692 m/s
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)
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.