Posted by sjeba on .
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

science 
Devron,
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. 
science 
Graham,
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. 
science 
Devron,
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.